Geometric Progression Example-1

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Home > Word Problems calculators > Geometric Progression Word Problems example

Geometric Progression examples ( Enter your problem )

( Enter your problem )

8. Arithmetic Progression
(Previous method)

1. Example-1

Problem : 1 / 23 [ Geometric Progression ] Enter your problem 1. For given geometric progression series 3,6,12,24,48 ,... find 10 th term and addition of first 10 th terms.


Solution:	Here a=3, $a = 3,$ r=63=2 $r = 6/3 = 2$ We know that, an=a×rn-1 $a_n = a × r^(n-1)$ a10=3×210-1 $a_10 = 3 × 2^(10 - 1)$ =3×512 $= 3 × 512$ =1536 $= 1536$ We know that, Sn=a⋅rn-1r-1 $S_n = a * (r^n - 1)/(r - 1)$ ∴S10=3×210-12-1 $:. S_10 = 3 × (2^10 - 1)/(2 - 1)$ ⇒S10=3×1024-11 $=> S_10 = 3 × (1024 - 1)/1$ ⇒S10=3×10231 $=> S_10 = 3 × 1023/1$ ⇒S10=3069 $=> S_10 = 3069$ Hence, 10th $10^(th)$ term of the given series is 1536 $1536$ and sum of first 10th $10^(th)$ term is 3069 $3069$

Problem : 2 / 23 [ Geometric Progression ] Enter your problem 2. For given geometric progression series 3,6,12,24,48 ,... then find n such that S(n) = 3069 .


Solution:	Here a=3, $a = 3,$ r=63=2 $r = 6/3 = 2$ We know that, Sn=a⋅rn-1r-1 $S_n = a * (r^n - 1)/(r - 1)$ ⇒Sn=3×(2)n-12-1 $=> S_n = 3 × ((2)^n - 1) / (2 - 1)$ ⇒3069=3×(2)n-11 $=> 3069 = 3 × ((2)^n - 1)/(1)$ ⇒2n-1=3069×13 $=> 2^n - 1 = 3069 × (1) / 3$ ⇒2n-1=1023 $=> 2^n - 1 = 1023$ ⇒2n=1024 $=> 2^n = 1024$ ⇒2n=210 $=> 2^n = 2^10$ ⇒n=10 $=> n = 10$

Problem : 3 / 23 [ Geometric Progression ] Enter your problem 3. For given geometric progression series 3,6,12,24,48 ,... then find n such that f(n) = 1536 .


Solution:	Here a=3, $a = 3,$ r=63=2 $r = 6/3 = 2$ Let n be the term such that f(n)=1536 $f(n) = 1536$ We know that, an=a×rn-1 $a_n = a × r^(n-1)$ ⇒3×2n-1=1536 $=> 3 × 2^(n-1) = 1536$ ⇒2n-1=512 $=> 2^(n-1) = 512$ ⇒2n-1=29 $=> 2^(n-1) = 2^9$ ⇒n-1=9 $=> n - 1 = 9$ ⇒n=9+1 $=> n = 9 + 1$ ⇒n=10 $=> n = 10$

Problem : 4 / 23 [ Geometric Progression ] Enter your problem 4. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 then find f( 3 ) and S( 3 ).


Solution:	We know that, an=a×rn-1 $a_n = a × r^(n-1)$ Here a1=2 $a_1 = 2$ ⇒a×r1-1=2 $=> a × r^(1 - 1) = 2$ ⇒a×r0=2 $=> a × r^0 = 2$ ⇒a=2→(1) $=> a = 2 ->(1)$ a4=54 $a_4 = 54$ ⇒a×r4-1=54 $=> a × r^(4 - 1) = 54$ ⇒a×r3=54→(2) $=> a × r^3 = 54 ->(2)$ Solving (1) $(1)$ and (2) $(2)$ , we get a=2 $a = 2$ and r=3 $r = 3$ We know that, an=a×rn-1 $a_n = a × r^(n-1)$ a3=2×33-1 $a_3 = 2 × 3^(3 - 1)$ =2×9 $= 2 × 9$ =18 $= 18$ We know that, Sn=a⋅rn-1r-1 $S_n = a * (r^n - 1)/(r - 1)$ ∴S3=2×33-13-1 $:. S_3 = 2 × (3^3 - 1)/(3 - 1)$ ⇒S3=2×27-12 $=> S_3 = 2 × (27 - 1)/2$ ⇒S3=2×262 $=> S_3 = 2 × 26/2$ ⇒S3=26 $=> S_3 = 26$

Problem : 5 / 23 [ Geometric Progression ] Enter your problem 5. For geometric progression f( 1 ) = 2 , f( 4 ) = 54 , then find n such that f(n) = 18 .


Solution:	We know that, an=a×rn-1 $a_n = a × r^(n-1)$ Here a1=2 $a_1 = 2$ ⇒a×r1-1=2 $=> a × r^(1 - 1) = 2$ ⇒a×r0=2 $=> a × r^0 = 2$ ⇒a=2→(1) $=> a = 2 ->(1)$ a4=54 $a_4 = 54$ ⇒a×r4-1=54 $=> a × r^(4 - 1) = 54$ ⇒a×r3=54→(2) $=> a × r^3 = 54 ->(2)$ Solving (1) $(1)$ and (2) $(2)$ , we get a=2 $a = 2$ and r=3 $r = 3$ Let n be the term such that f(n)=18 $f(n) = 18$ We know that, an=a×rn-1 $a_n = a × r^(n-1)$ ⇒2×3n-1=18 $=> 2 × 3^(n-1) = 18$ ⇒3n-1=9 $=> 3^(n-1) = 9$ ⇒3n-1=32 $=> 3^(n-1) = 3^2$ ⇒n-1=2 $=> n - 1 = 2$ ⇒n=2+1 $=> n = 2 + 1$ ⇒n=3 $=> n = 3$

Problem : 6 / 23 [ Geometric Progression ] Enter your problem 6. For geometric progression addition of 3 terms is 26 and their multiplication is 216 , then that numbers


Solution:	Let the terms are (ar),a,(a×r) $(a / r) , a , (a × r)$ Addition : (ar)+a+(a×r)=26→(1) $(a / r) + a + (a × r) = 26 ->(1)$ Multiplication : (ar)×a×(a×r)=216 $(a / r) × a × (a × r) = 216$ ⇒a3=216 $=> a^3 = 216$ ⇒a=6 $=> a = 6$ Now putting a=6 $a = 6$ in (1) $(1)$ , we get (6r)+6+(6×r)=26 $(6 / r) + 6 + (6 × r) = 26$ (6r)+(6×r)=20 $(6 / r) + (6 × r) = 20$ ⇒r=3 $=> r = 3$ or r=13 $r = 1/3$ Now, a=6 $a = 6$ and r=3⇒(63),6,(6×3)⇒2,6,18 $r = 3 => (6 / 3) , 6 , (6 × 3) => 2 , 6 , 18$ or a=6 $a = 6$ and r=13⇒(613),6,(6×(13))⇒18,6,2 $r = 1/3 => (6 / (1/3)) , 6 , (6 × (1/3)) => 18 , 6 , 2$

Problem : 7 / 23 [ Geometric Progression ] Enter your problem 7. For geometric progression multiplication of 5 terms is 1 and 5 th term is 81 times then the 1 th term.


Solution:	Let the terms are (ar2),(ar),a,(a×r),(a×r2) $(a / r^2) , (a / r) , a , (a × r) , (a × r^2)$ Multiplication : (ar2)⋅(ar)⋅a⋅(a×r)⋅(a×r2)=1 $(a / r^2) * (a / r) * a * (a × r) * (a × r^2) = 1$ ⇒a5=1 $=> a^5 = 1$ ⇒a=1 $=> a = 1$ Now 5th $5^(th)$ term is 81 $81$ times than the 1st $1^(st)$ term ⇒(a×r2)=81(ar2) $=> (a × r^2) = 81 (a / r^2)$ ⇒r4=81 $=> r^4 = 81$ ⇒r=3 $=> r = 3$ Now, a=1 $a = 1$ and r=3⇒(132),(13),1,(1×3),(1×32)⇒0.1111,0.3333,1,3,9 $r = 3 => (1 / 3^2) , (1 / 3) , 1 , (1 × 3) , (1 × 3^2) => 0.1111 , 0.3333 , 1 , 3 , 9$

Problem : 8 / 23 [ Geometric Progression ] Enter your problem 8. Arithmetic mean of two number is 13 and geometric mean is 12 , then find that numbers


Solution:	Let the two terms be a $a$ and b $b$ . ⇒A=a+b2=13 $=> A = (a + b)/2 = 13$ and G=√ab=12 $G = sqrt(ab) = 12$ ⇒a+b=26→(1) $=> a + b = 26 ->(1)$ and ab=144→(2) $ab = 144 ->(2)$ ⇒b=26-a→(3) $=> b = 26 - a ->(3)$ Now putting the value of (3) $(3)$ in (2) $(2)$ , we get a(26-a)=144 $a(26 - a) = 144$ ⇒26a-a2=144 $=> 26a - a^2 = 144$ ⇒a2-26a+144=0 $=> a^2 - 26a + 144 = 0$ ⇒(a-8)(a-18)=0 $=> (a - 8)(a - 18) = 0$ ⇒a=8 $=> a = 8$ or a=18 $a = 18$ ⇒a=8⇒b=26-a=26-8=18 $=> a = 8 => b = 26 - a = 26 - 8 = 18$ and a=18⇒b=26-a=26-18=8 $a = 18 => b = 26 - a = 26 - 18 = 8$ ∴ $:.$ Required numbers are 8 $8$ and 18 $18$

Problem : 9 / 23 [ Geometric Progression ] Enter your problem 9. Two numbers are in the ratio 9 : 16 and difference of arithmetic mean and geometric mean is 1 , then find that numbers


Solution:	Here ratio of the two terms is 9:16 $9:16$ . Let the two terms be 9x $9x$ and 16x $16x$ . Now difference of their Arithmetic mean A $A$ and Geometric mean G $G$ is 1 $1$ . ⇒A-G=1 $=> A - G = 1$ (Because A $A$ > G $G$ ). ⇒9x+16x2-√9x×16x=1 $=> (9x + 16x)/2 - sqrt(9x × 16x) = 1$ . ⇒25x2-√144x=1 $=> (25x)/2 - sqrt(144) x = 1$ . ⇒x=2 $=> x = 2$ ⇒ $=>$ The required two terms are 9×2 $9 × 2$ and 16×2 $16 × 2$ . ⇒ $=>$ The required two terms are 18 $18$ and 32 $32$ .

Problem : 10 / 23 [ Geometric Progression ] Enter your problem 10. Find 6 arithmetic mean between 3 and 24 .


Solution:	Let A1,A2,A3,A4,A5,A6 $A_1, A_2, A_3, A_4, A_5, A_6$ be the 6 $6$ arithmetic mean between 3 $3$ and 24 $24$ . ∴3,A1,A2,A3,A4,A5,A6,24 $:. 3, A_1, A_2, A_3, A_4, A_5, A_6, 24$ are in AP . Here First term a=3 $a = 3$ and Last term b=24 $b = 24$ Comman difference d=b-an+1=24-36+1=217=3 $d = (b-a)/(n+1) = (24-3)/(6+1) = 21/7 = 3$ ∴ $:.$ Required Arithemetic Means, A1=T2=a+d=3+3=6 $A_1 = T_2 = a + d = 3 + 3 = 6$ A2=T3=a+2d=3+2(3)=9 $A_2 = T_3 = a + 2d = 3 + 2(3) = 9$ A3=T4=a+3d=3+3(3)=12 $A_3 = T_4 = a + 3d = 3 + 3(3) = 12$ A4=T5=a+4d=3+4(3)=15 $A_4 = T_5 = a + 4d = 3 + 4(3) = 15$ A5=T6=a+5d=3+5(3)=18 $A_5 = T_6 = a + 5d = 3 + 5(3) = 18$ A6=T7=a+6d=3+6(3)=21 $A_6 = T_7 = a + 6d = 3 + 6(3) = 21$

Problem : 11 / 23 [ Geometric Progression ] Enter your problem 11. Find 3 geometric mean between 1 and 256 .


Solution:	Let a1,a2,a3 $a_1, a_2, a_3$ be the 3 $3$ geometric mean between 1 $1$ and 256 $256$ . ∴1,a1,a2,a3,256 $:. 1, a_1, a_2, a_3, 256$ are in GP . Here First term a=1 $a = 1$ and Last term b=256 $b = 256$ Comman ratio r=(ba)1n+1=(2561)14=4 $r = (b/a)^(1/(n+1)) = (256/1)^(1/4) = 4$ ∴ $:.$ Required Geometic Means, a1=a⋅r=1⋅4=4 $a_1 = a * r = 1 * 4 = 4$ a2=ar2=1×42=16 $a_2 = a r^2 = 1 × 4^2 = 16$ a3=ar3=1×43=64 $a_3 = a r^3 = 1 × 4^3 = 64$

Problem : 12 / 23 [ Geometric Progression ] Enter your problem 12. Prove that 1+(1+2)+(1+2+3)+...n $1 + (1 + 2) + (1 + 2 + 3) + ... n$ terms =n6(n+1)(n+2) $= n/6 (n + 1) (n + 2)$


Solution:	L.H.S. =1+(1+2)+(1+2+3)+...+n $= 1 + (1 + 2) + (1 + 2 + 3) + ... + n$ terms =∑[f(n)] $= sum [ f(n) ]$ =∑[∑n] $= sum [ sum n ]$ =∑[n2(n+1)] $= sum [ n/2 (n + 1) ]$ =12∑n2+12∑n $= 1/2 sum n^2 + 1/2 sum n$ =12⋅n(n+1)(2n+1)6+12⋅n(n+1)2 $= 1/2 * (n (n + 1) (2n + 1))/6 + 1/2 * (n (n + 1))/2$ =n(n+1)(2n+1)12+n(n+1)4 $= (n (n + 1) (2n + 1))/12 + (n (n + 1))/4$ =n(n+1)(2n+1+3)12 $= (n (n + 1) (2n + 1 + 3))/12$ =n(n+1)(2n+4)12 $= (n (n + 1) (2n + 4))/12$ =n(n+1)(n+2)6 $= (n (n + 1) (n + 2))/6$ = $=$ R.H.S. (Proved)

Problem : 13 / 23 [ Geometric Progression ] Enter your problem 13. Prove that 1⋅(22-32)+2⋅(32-42)+3⋅(42-52)+...n $1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n$ terms =n6(n+1)(4n+11) $= n/6 (n + 1) (4n + 11)$


Solution:	L.H.S. =1×(22-32)+2×(32-42)+3×(42-52)+...n $= 1 × (2^2 - 3^2) + 2 × (3^2 - 4^2) + 3 × (4^2 - 5^2) + ... n$ terms =∑[f(n)] $= sum [ f(n) ]$ =∑[n((n+1)2-(n+2)2)] $= sum [ n ((n + 1)^2 - (n + 2)^2) ]$ =∑[n(n2+2n+1-n2-2n-4)] $= sum [ n (n^2 + 2n + 1 - n^2 - 2n - 4) ]$ =∑[n(-2n-3)] $= sum [ n (-2n - 3) ]$ =∑(-2n2-3n)] $= sum (-2n^2 - 3n) ]$ =-2∑n2-3∑n $= -2 sum n^2 - 3 sum n$ =-2⋅n(n+1)(2n+1)6-3⋅n(n+1)2 $= -2 * (n (n + 1) (2n + 1))/6 - 3 * (n (n + 1))/2$ =-n6(n+1)[2(2n+1)+9] $= - n/6 (n + 1) [ 2(2n + 1) + 9 ]$ =-n6(n+1)(4n+11) $= - n/6 (n + 1) (4n + 11)$ = $=$ R.H.S. (Proved)

Problem : 14 / 23 [ Geometric Progression ] Enter your problem 14. 1+x4+32+4+x6+62+7+x8+92+...3n $1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n$ terms


Solution:	S3n=1+x4+32+4+x6+62+7+x8+92+...3n $S_(3n) = 1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n$ terms =(1+4+7+...n terms)+(x4+x6+x8+...n terms)+(32+62+92+...n terms) $= (1 + 4 + 7 + ... n " terms") + (x^4 + x^6 + x^8 + ... n " terms") + (3^2 + 6^2 + 9^2+ ... n " terms")$ =(1+4+7+...n terms)+x4(1+x2+x4+...n terms)+32(1+22+32+...n terms) $= (1 + 4 + 7 + ... n " terms") + x^4 (1 + x^2 + x^4 + ... n " terms") + 3^2 (1 + 2^2 + 3^2 + ... n " terms")$ =n2[2×1+(n-1)×3]+x4(x2)n-1x2-1+9×n6(n+1)(2n+1) $= n/2 [ 2 × 1 + (n - 1) × 3 ] + x^4 [ (x^2)^n - 1 ] / ( x^2 - 1) + 9 × n/6 (n + 1) (2n + 1)$ =n2(3n-1)+x4x2n-1x2-1+3n2(n+1)(2n+1) $= n/2 (3n - 1) + x^4 (x^(2n) - 1) / ( x^2 - 1) + (3n)/2 (n + 1) (2n + 1)$

Problem : 15 / 23 [ Geometric Progression ] Enter your problem 15. 1 + (1 + 3) + (1 + 3 + 5) + ... n terms


Solution:	Sn=∑[f(n)] $S_n = sum [ f(n) ]$ =∑[∑(2n-1)] $= sum [ sum (2n - 1) ]$ =∑[2∑n-∑1] $= sum [ 2 sum n - sum 1 ]$ =∑[2⋅n(n+1)2-n] $= sum [ 2 * (n (n + 1))/2 - n ]$ =∑[n2+n-n] $= sum [ n^2 + n - n ]$ =∑n2 $= sum n^2$ =n(n+1)(2n+1)6 $= (n (n + 1) (2n + 1))/6$

Problem : 16 / 23 [ Geometric Progression ] Enter your problem 16. Prove that for all n belongs to N, 12⋅n+22⋅(n-1)+32⋅(n-2)+...+n2⋅1=n12(n+1)2(n+2) $1^2 * n + 2^2 * (n - 1) + 3^2 * (n - 2) + ... + n^2 * 1 = n/12 (n + 1)^2 (n + 2)$


Solution:	Here, f(r)=r2(n-r+1)=(n+1)r2-r3 $f(r) = r^2 (n -r + 1) = (n + 1) r^2 - r^3$ Now, L.H.S. =12×n+22(n-1)+32(n-2)+...+n2×1 $= 1^2 × n + 2^2 (n - 1) + 3^2 (n - 2) + ... + n^2 × 1$ =∑[(n+1)r2-r3] $= sum [ (n + 1) r^2 - r^3 ]$ (Where r = 1 to n) =(n+1)∑r2-∑r3 $= (n + 1) sum r^2 - sum r^3$ (Where r = 1 to n) =(n+1)⋅n(n+1)(2n+1)6-n2(n+1)24 $= (n + 1) * (n (n + 1) (2n + 1))/6 - (n^2 (n + 1)^2)/4$ =n(n+1)2[2(2n+1)-3n]12 $= (n (n + 1)^2 [ 2 (2n + 1) - 3 n])/12$ =n(n+1)2(n+2)12 $= (n (n + 1)^2 (n + 2))/12$ = $=$ R.H.S. (Proved)

Problem : 17 / 23 [ Geometric Progression ] Enter your problem 17. Prove that 1⋅22+3⋅52+5⋅82+...n $1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n$ terms =n2(9n3+4n2-4n-1) $= n/2 (9n^3 + 4n^2 - 4n - 1)$


Solution:	L.H.S. =1×22+3×52+5×82+...n $= 1 × 2^2 + 3 × 5^2 + 5 × 8^2 + ... n$ terms =∑[f(n)] $= sum [ f(n) ]$ =∑[(2n-1)(3n-1)2] $= sum [ (2n - 1)(3n - 1)^2 ]$ =∑[(2n-1)(9n2-6n+1)] $= sum [ (2n - 1)(9n^2 - 6n + 1)]$ =∑[18n3-21n2+8n-1] $= sum [ 18n^3 - 21n^2 + 8n - 1]$ =18∑n3-21∑n2+8∑n-∑1 $= 18 sum n^3 - 21 sum n^2 + 8 sum n - sum 1$ =18⋅n2(n+1)24-21⋅n(n+1)(2n+1)6+8⋅n(n+1)2-n $= 18 * (n^2 (n+1)^2)/4 - 21 * (n (n + 1) (2n + 1))/6 + 8 * (n (n+1))/2 - n$ =n2[9n(n+1)2-7(n+1)(2n+1)+8(n+1)-2] $= n/2 [ 9 n (n + 1)^2 - 7(n + 1) (2n + 1) + 8 (n+1) - 2 ]$ =n2[9n(n2+2n+1)-7(2n2+3n+1)+8n+8-2] $= n/2 [ 9 n (n^2 + 2n +1) - 7 (2n^2 + 3n + 1) + 8 n + 8 - 2 ]$ =n2[9n3+18n2+9n-14n2-21n-7+8n+8-2] $= n/2 [ 9 n^3 + 18 n^2 + 9n - 14n^2 - 21n - 7 + 8 n + 8 - 2 ]$ =n2[9n3+4n2-4n-1] $= n/2 [ 9 n^3 + 4 n^2 - 4n - 1 ]$ = $=$ R.H.S. (Proved)

Problem : 18 / 23 [ Geometric Progression ] Enter your problem 18. Prove that 12+(12+22)+(12+22+32)+...n $1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n$ terms =n12(n+1)2(n+2) $= n/12 (n + 1)^2 (n + 2)$


Solution:	L.H.S. =12+(12+22)+(12+22+32)+...n $= 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n$ terms =∑[f(n)] $= sum [ f(n) ]$ =∑[∑n2] $= sum [ sum n^2 ]$ =∑[n6(n+1)(2n+1)] $= sum [ n/6 (n + 1) (2n + 1) ]$ =∑[n6(2n2+3n+1)] $= sum [ n/6 (2n^2 + 3n + 1) ]$ =∑[16(2n3+3n2+n)] $= sum [ 1/6 (2n^3 + 3n^2 + n) ]$ =16[2∑n3+3∑n2+∑n] $= 1/6 [ 2 sum n^3 + 3 sum n^2 + sum n ]$ =16[2⋅n2(n+1)24+3⋅n(n+1)(2n+1)6+n(n+1)2] $= 1/6 [ 2 * (n^2 (n+1)^2)/4 + 3 * (n (n + 1) (2n + 1))/6 + (n (n + 1))/2 ]$ =16[n2(n+1)22+n(n+1)(2n+1)2+n(n+1)2] $= 1/6 [ (n^2 (n+1)^2)/2 + (n (n + 1) (2n + 1))/2 + (n (n + 1))/2 ]$ =16⋅n(n+1)2[n(n+1)+(2n+1)+1] $= 1/6 * (n (n + 1))/2 [ n (n+1) + (2n + 1) + 1 ]$ =n12(n+1)[n2+n+2n+2] $= n/12 (n + 1) [ n^2 + n + 2n + 2 ]$ =n12(n+1)[n2+3n+2] $= n/12 (n + 1) [ n^2 + 3n + 2 ]$ =n12(n+1)[(n+2)(n+1)] $= n/12 (n + 1) [ (n + 2) (n + 1) ]$ =n12(n+1)2(n+2) $= n/12 (n + 1)^2 (n + 2)$ = $=$ R.H.S. (Proved)

Problem : 19 / 23 [ Geometric Progression ] Enter your problem 19. Prove that 2 + 5 + 10 + 17 + ... n terms = n6(2n2+3n+7) $n/6 (2n^2 + 3n + 7)$


Solution:	L.H.S. =2+5+10+17+...n $= 2 + 5 + 10 + 17 + ... n$ terms =∑[1+n2] $= sum [ 1 + n^2 ]$ =∑1+∑n2 $= sum 1 + sum n^2$ =n+n(n+1)(2n+1)6 $= n + (n (n + 1) (2n + 1))/6$ =n6[6+(2n2+3n+1)] $= n/6 [ 6 + (2n^2 + 3n + 1) ]$ =n6(2n2+3n+7) $= n/6 (2n^2 + 3n + 7)$ = $=$ R.H.S. (Proved)

Problem : 20 / 23 [ Geometric Progression ] Enter your problem 20. Prove that ∑[∑(2n-3)]=n6(n+1)(2n-5) $sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)$


Solution:	L.H.S. =∑[∑(2n-3)] $= sum [ sum (2n -3) ]$ =∑[∑2n-∑3] $= sum [ sum 2n - sum 3 ]$ =∑[2×n2(n+1)-3n] $= sum [ 2 × n/2 (n + 1) -3n ]$ =∑[n(n+1)-3n] $= sum [ n (n + 1) - 3n ]$ =∑[n2+n-3n] $= sum [ n^2 + n - 3n ]$ =∑[n2-2n] $= sum [ n^2 - 2n ]$ =∑n2-2∑n $= sum n^2 - 2 sum n$ =n(n+1)(2n+1)6-2⋅n(n+1)2 $= (n (n + 1) (2n + 1))/6 - 2 * (n (n + 1))/2$ =n(n+1)[(2n+1)-6]6 $= (n (n + 1) [ (2n + 1) - 6 ])/6$ =n(n+1)(2n-5)6 $= (n (n + 1) (2n - 5))/6$ = $=$ R.H.S. (Proved)

Problem : 21 / 23 [ Geometric Progression ] Enter your problem 21. For geometric progression, find 1+1√2+12+12⋅√2+...10 $1 + 1/sqrt(2) + 1/2 + 1/(2sqrt(2)) + ... 10$ terms ( For geometric progression, find 1+1√x+1x+1x⋅√x+...n $1 + 1/sqrt(x) + 1/x + 1/(xsqrt(x)) + ... n$ terms where x = 2 and n = 10 . )


Solution:	Here a=1,r=1√2,n=10 $a = 1, r = 1 / sqrt(2), n = 10$ We know that, Sn=a⋅1-rn1-r $S_n = a * (1 - r^n)/(1 - r)$ ∴S10=1×1-(1√2)101-1√2 $:. S_10 = 1 × (1 - (1 / sqrt(2))^10) / (1 - 1 / sqrt(2))$ =1×1-(12)5√2-1√2 $= 1 × (1 - (1 / 2)^5) / ((sqrt(2) - 1) / sqrt(2))$ =√2×1×1-132√2-1 $= sqrt(2) × 1 × (1 - 1 / 32) / (sqrt(2) - 1)$ =√2×1×32-132√2-1×√2+1√2+1 $= sqrt(2) × 1 × ((32 - 1) / 32) / (sqrt(2) - 1) × (sqrt(2) + 1) / (sqrt(2) + 1)$ =√2×1×31322-1×(√2+1) $= sqrt(2) × 1 × (31/32) / (2 - 1) × (sqrt(2) + 1)$ =(3132)×√2×(√2+1) $= (31/32) × sqrt(2) × (sqrt(2) + 1)$ =(3132)×(2+√2) $= (31/32) × (2 + sqrt(2))$

Problem : 22 / 23 [ Geometric Progression ] Enter your problem 22. Find 12+22+...+102 $1^2 + 2^2 + ...+ 10^2$ , ( Find a2+b2+...+n2 $a^2 + b^2 + ... + n^2$ , where a = 1 , b = 2 and n = 10 . )


Solution:	Here Find 12+22+32+...+102 $1^2 + 2^2 + 3^2 + ... + 10^2$ =∑i2 $= sum i^2$ (where i=1,...,10 $i = 1,...,10$ ) =n(n+1)(2n+1)6 $= (n (n + 1)(2n + 1)) / 6$ (where n=10 $n = 10$ ) =10(10+1)(2×10+1)6 $= (10 (10 + 1) (2×10 + 1)) / 6$ =10×11×216 $= (10 × 11 × 21) / 6$ =5×11×7 $= 5 × 11 × 7$ =385 $= 385$

Problem : 23 / 23 [ Geometric Progression ] Enter your problem 23. Find 12+22+...10 $1^2 + 2^2 + ... 10$ terms, ( Find a2+b2+...n $a^2 + b^2 + ... n$ terms, where a = 1 , b = 2 and n = 10 . )


Solution:	Here Find 12+22+32+...10terms $1^2 + 2^2 + 3^2 + ... 10 "terms"$ =∑i2 $= sum i^2$ (where i=1,...,10 $i = 1,...,10$ ) =n(n+1)(2n+1)6 $= (n (n + 1)(2n + 1)) / 6$ (where n=10 $n = 10$ ) =10(10+1)(2×10+1)6 $= (10 (10 + 1) (2×10 + 1)) / 6$ =10×11×216 $= (10 × 11 × 21) / 6$ =5×11×7 $= 5 × 11 × 7$ =385 $= 385$