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.
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| Solution:
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Here `a = 3,`
`r = 6/3 = 2`
We know that, `a_n = a × r^(n-1)`
`a_10 = 3 × 2^(10 - 1)`
`= 3 × 512`
`= 1536`
We know that, `S_n = a * (r^n - 1)/(r - 1)`
`:. S_10 = 3 × (2^10 - 1)/(2 - 1)`
`=> S_10 = 3 × (1024 - 1)/1`
`=> S_10 = 3 × 1023/1`
`=> S_10 = 3069`
Hence, `10^(th)` term of the given series is `1536` and sum of first `10^(th)` term is `3069`
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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 .
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| Solution:
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Here `a = 3,`
`r = 6/3 = 2`
We know that, `S_n = a * (r^n - 1)/(r - 1)`
`=> S_n = 3 × ((2)^n - 1) / (2 - 1)`
`=> 3069 = 3 × ((2)^n - 1)/(1)`
`=> 2^n - 1 = 3069 × (1) / 3`
`=> 2^n - 1 = 1023`
`=> 2^n = 1024`
`=> 2^n = 2^10`
`=> n = 10`
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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 .
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| Solution:
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Here `a = 3,`
`r = 6/3 = 2`
Let n be the term such that `f(n) = 1536`
We know that, `a_n = a × r^(n-1)`
`=> 3 × 2^(n-1) = 1536`
`=> 2^(n-1) = 512`
`=> 2^(n-1) = 2^9`
`=> n - 1 = 9`
`=> n = 9 + 1`
`=> n = 10`
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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 ).
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| Solution:
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We know that, `a_n = a × r^(n-1)`
Here `a_1 = 2`
`=> a × r^(1 - 1) = 2`
`=> a × r^0 = 2`
`=> a = 2 ->(1)`
`a_4 = 54`
`=> a × r^(4 - 1) = 54`
`=> a × r^3 = 54 ->(2)`
Solving `(1)` and `(2)`, we get `a = 2` and `r = 3`
We know that, `a_n = a × r^(n-1)`
`a_3 = 2 × 3^(3 - 1)`
`= 2 × 9`
`= 18`
We know that, `S_n = a * (r^n - 1)/(r - 1)`
`:. S_3 = 2 × (3^3 - 1)/(3 - 1)`
`=> S_3 = 2 × (27 - 1)/2`
`=> S_3 = 2 × 26/2`
`=> S_3 = 26`
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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 .
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| Solution:
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We know that, `a_n = a × r^(n-1)`
Here `a_1 = 2`
`=> a × r^(1 - 1) = 2`
`=> a × r^0 = 2`
`=> a = 2 ->(1)`
`a_4 = 54`
`=> a × r^(4 - 1) = 54`
`=> a × r^3 = 54 ->(2)`
Solving `(1)` and `(2)`, we get `a = 2` and `r = 3`
Let n be the term such that `f(n) = 18`
We know that, `a_n = a × r^(n-1)`
`=> 2 × 3^(n-1) = 18`
`=> 3^(n-1) = 9`
`=> 3^(n-1) = 3^2`
`=> n - 1 = 2`
`=> n = 2 + 1`
`=> n = 3`
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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
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| Solution:
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Let the terms are `(a / r) , a , (a × r)`
Addition : `(a / r) + a + (a × r) = 26 ->(1)`
Multiplication : `(a / r) × a × (a × r) = 216`
`=> a^3 = 216`
`=> a = 6`
Now putting `a = 6` in `(1)`, we get
`(6 / r) + 6 + (6 × r) = 26`
`(6 / r) + (6 × r) = 20`
`=> r = 3` or `r = 1/3`
Now, `a = 6` and `r = 3 => (6 / 3) , 6 , (6 × 3) => 2 , 6 , 18`
or `a = 6` and `r = 1/3 => (6 / (1/3)) , 6 , (6 × (1/3)) => 18 , 6 , 2`
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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.
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| Solution:
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Let the terms are `(a / r^2) , (a / r) , a , (a × r) , (a × r^2)`
Multiplication : `(a / r^2) * (a / r) * a * (a × r) * (a × r^2) = 1`
`=> a^5 = 1`
`=> a = 1`
Now `5^(th)` term is `81` times than the `1^(st)` term
`=> (a × r^2) = 81 (a / r^2)`
`=> r^4 = 81`
`=> r = 3`
Now, `a = 1` and `r = 3 => (1 / 3^2) , (1 / 3) , 1 , (1 × 3) , (1 × 3^2) => 0.1111 , 0.3333 , 1 , 3 , 9`
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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
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| Solution:
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Let the two terms be `a` and `b`.
`=> A = (a + b)/2 = 13` and `G = sqrt(ab) = 12`
`=> a + b = 26 ->(1)` and `ab = 144 ->(2)`
`=> b = 26 - a ->(3)`
Now putting the value of `(3)` in `(2)`, we get
`a(26 - a) = 144`
`=> 26a - a^2 = 144`
`=> a^2 - 26a + 144 = 0`
`=> (a - 8)(a - 18) = 0`
`=> a = 8` or `a = 18`
`=> a = 8 => b = 26 - a = 26 - 8 = 18`
and `a = 18 => b = 26 - a = 26 - 18 = 8`
`:.` Required numbers are `8` and `18`
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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
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| Solution:
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Here ratio of the two terms is `9:16`.
Let the two terms be `9x` and `16x`.
Now difference of their Arithmetic mean `A` and Geometric mean `G` is `1`.
`=> A - G = 1` (Because `A` > `G`).
`=> (9x + 16x)/2 - sqrt(9x × 16x) = 1`.
`=> (25x)/2 - sqrt(144) x = 1`.
`=> x = 2`
`=>` The required two terms are `9 × 2` and `16 × 2`.
`=>` The required two terms are `18` and `32`.
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Problem : 10 / 23
[ Geometric Progression ]
Enter your problem
10.
Find 6 arithmetic mean between 3 and 24 .
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| Solution:
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Let ` A_1, A_2, A_3, A_4, A_5, A_6` be the `6` arithmetic mean between `3` and `24`.
`:. 3, A_1, A_2, A_3, A_4, A_5, A_6, 24` are in AP .
Here First term `a = 3` and Last term `b = 24`
Comman difference `d = (b-a)/(n+1) = (24-3)/(6+1) = 21/7 = 3`
`:.` Required Arithemetic Means,
`A_1 = T_2 = a + d = 3 + 3 = 6`
`A_2 = T_3 = a + 2d = 3 + 2(3) = 9`
`A_3 = T_4 = a + 3d = 3 + 3(3) = 12`
`A_4 = T_5 = a + 4d = 3 + 4(3) = 15`
`A_5 = T_6 = a + 5d = 3 + 5(3) = 18`
`A_6 = T_7 = a + 6d = 3 + 6(3) = 21`
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Problem : 11 / 23
[ Geometric Progression ]
Enter your problem
11.
Find 3 geometric mean between 1 and 256 .
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| Solution:
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Let ` a_1, a_2, a_3` be the `3` geometric mean between `1` and `256`.
`:. 1, a_1, a_2, a_3, 256` are in GP .
Here First term `a = 1` and Last term `b = 256`
Comman ratio `r = (b/a)^(1/(n+1)) = (256/1)^(1/4) = 4`
`:.` Required Geometic Means,
`a_1 = a * r = 1 * 4 = 4`
`a_2 = a r^2 = 1 × 4^2 = 16`
`a_3 = a r^3 = 1 × 4^3 = 64`
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Problem : 12 / 23
[ Geometric Progression ]
Enter your problem
12.
Prove that `1 + (1 + 2) + (1 + 2 + 3) + ... n` terms `= n/6 (n + 1) (n + 2)`
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| Solution:
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L.H.S. `= 1 + (1 + 2) + (1 + 2 + 3) + ... + n` terms
`= sum [ f(n) ]`
`= sum [ sum n ]`
`= sum [ n/2 (n + 1) ]`
`= 1/2 sum n^2 + 1/2 sum n`
`= 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 + 3))/12`
`= (n (n + 1) (2n + 4))/12`
`= (n (n + 1) (n + 2))/6`
`=` R.H.S. (Proved)
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Problem : 13 / 23
[ Geometric Progression ]
Enter your problem
13.
Prove that `1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n` terms `= n/6 (n + 1) (4n + 11)`
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| Solution:
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L.H.S. `= 1 × (2^2 - 3^2) + 2 × (3^2 - 4^2) + 3 × (4^2 - 5^2) + ... n` terms
`= sum [ f(n) ]`
`= sum [ n ((n + 1)^2 - (n + 2)^2) ]`
`= sum [ n (n^2 + 2n + 1 - n^2 - 2n - 4) ]`
`= sum [ n (-2n - 3) ]`
`= sum (-2n^2 - 3n) ]`
`= -2 sum n^2 - 3 sum n`
`= -2 * (n (n + 1) (2n + 1))/6 - 3 * (n (n + 1))/2`
`= - n/6 (n + 1) [ 2(2n + 1) + 9 ]`
`= - n/6 (n + 1) (4n + 11)`
`=` R.H.S. (Proved)
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Problem : 14 / 23
[ Geometric Progression ]
Enter your problem
14.
`1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n` terms
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| Solution:
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`S_(3n) = 1 + x^4 + 3^2 + 4 + x^6 + 6^2 + 7 + x^8 + 9^2 + ... 3n` 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") + x^4 (1 + x^2 + x^4 + ... n " terms") + 3^2 (1 + 2^2 + 3^2 + ... n " terms")`
`= n/2 [ 2 × 1 + (n - 1) × 3 ] + x^4 [ (x^2)^n - 1 ] / ( x^2 - 1) + 9 × n/6 (n + 1) (2n + 1)`
`= n/2 (3n - 1) + x^4 (x^(2n) - 1) / ( x^2 - 1) + (3n)/2 (n + 1) (2n + 1)`
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Problem : 15 / 23
[ Geometric Progression ]
Enter your problem
15.
1 + (1 + 3) + (1 + 3 + 5) + ... n terms
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| Solution:
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`S_n = sum [ f(n) ]`
`= sum [ sum (2n - 1) ]`
`= sum [ 2 sum n - sum 1 ]`
`= sum [ 2 * (n (n + 1))/2 - n ]`
`= sum [ n^2 + n - n ]`
`= sum n^2`
`= (n (n + 1) (2n + 1))/6`
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Problem : 16 / 23
[ Geometric Progression ]
Enter your problem
16.
Prove that for all n belongs to N, `1^2 * n + 2^2 * (n - 1) + 3^2 * (n - 2) + ... + n^2 * 1 = n/12 (n + 1)^2 (n + 2)`
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| Solution:
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Here, `f(r) = r^2 (n -r + 1) = (n + 1) r^2 - r^3`
Now, L.H.S. `= 1^2 × n + 2^2 (n - 1) + 3^2 (n - 2) + ... + n^2 × 1`
`= sum [ (n + 1) r^2 - r^3 ]` (Where r = 1 to n)
`= (n + 1) sum r^2 - sum r^3` (Where r = 1 to n)
`= (n + 1) * (n (n + 1) (2n + 1))/6 - (n^2 (n + 1)^2)/4`
`= (n (n + 1)^2 [ 2 (2n + 1) - 3 n])/12`
`= (n (n + 1)^2 (n + 2))/12`
`=` R.H.S. (Proved)
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Problem : 17 / 23
[ Geometric Progression ]
Enter your problem
17.
Prove that `1 * 2^2 + 3 * 5^2 + 5 * 8^2 + ... n` terms `= n/2 (9n^3 + 4n^2 - 4n - 1)`
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| Solution:
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L.H.S. `= 1 × 2^2 + 3 × 5^2 + 5 × 8^2 + ... n` terms
`= sum [ f(n) ]`
`= sum [ (2n - 1)(3n - 1)^2 ]`
`= sum [ (2n - 1)(9n^2 - 6n + 1)]`
`= sum [ 18n^3 - 21n^2 + 8n - 1]`
`= 18 sum n^3 - 21 sum n^2 + 8 sum n - sum 1`
`= 18 * (n^2 (n+1)^2)/4 - 21 * (n (n + 1) (2n + 1))/6 + 8 * (n (n+1))/2 - n`
`= n/2 [ 9 n (n + 1)^2 - 7(n + 1) (2n + 1) + 8 (n+1) - 2 ]`
`= n/2 [ 9 n (n^2 + 2n +1) - 7 (2n^2 + 3n + 1) + 8 n + 8 - 2 ]`
`= n/2 [ 9 n^3 + 18 n^2 + 9n - 14n^2 - 21n - 7 + 8 n + 8 - 2 ]`
`= n/2 [ 9 n^3 + 4 n^2 - 4n - 1 ]`
`=` R.H.S. (Proved)
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Problem : 18 / 23
[ Geometric Progression ]
Enter your problem
18.
Prove that `1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n` terms `= n/12 (n + 1)^2 (n + 2)`
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| Solution:
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L.H.S. `= 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ... n` terms
`= sum [ f(n) ]`
`= sum [ sum n^2 ]`
`= sum [ n/6 (n + 1) (2n + 1) ]`
`= sum [ n/6 (2n^2 + 3n + 1) ]`
`= sum [ 1/6 (2n^3 + 3n^2 + n) ]`
`= 1/6 [ 2 sum n^3 + 3 sum n^2 + sum n ]`
`= 1/6 [ 2 * (n^2 (n+1)^2)/4 + 3 * (n (n + 1) (2n + 1))/6 + (n (n + 1))/2 ]`
`= 1/6 [ (n^2 (n+1)^2)/2 + (n (n + 1) (2n + 1))/2 + (n (n + 1))/2 ]`
`= 1/6 * (n (n + 1))/2 [ n (n+1) + (2n + 1) + 1 ]`
`= n/12 (n + 1) [ n^2 + n + 2n + 2 ]`
`= n/12 (n + 1) [ n^2 + 3n + 2 ]`
`= n/12 (n + 1) [ (n + 2) (n + 1) ]`
`= n/12 (n + 1)^2 (n + 2)`
`=` R.H.S. (Proved)
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Problem : 19 / 23
[ Geometric Progression ]
Enter your problem
19.
Prove that 2 + 5 + 10 + 17 + ... n terms = `n/6 (2n^2 + 3n + 7)`
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| Solution:
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L.H.S. `= 2 + 5 + 10 + 17 + ... n` terms
`= sum [ 1 + n^2 ]`
`= sum 1 + sum n^2`
`= n + (n (n + 1) (2n + 1))/6`
`= n/6 [ 6 + (2n^2 + 3n + 1) ]`
`= n/6 (2n^2 + 3n + 7)`
`=` R.H.S. (Proved)
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Problem : 20 / 23
[ Geometric Progression ]
Enter your problem
20.
Prove that `sum [ sum (2n -3) ] = n/6 (n + 1)(2n - 5)`
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| Solution:
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L.H.S. `= sum [ sum (2n -3) ]`
`= sum [ sum 2n - sum 3 ]`
`= sum [ 2 × n/2 (n + 1) -3n ]`
`= sum [ n (n + 1) - 3n ]`
`= sum [ n^2 + n - 3n ]`
`= sum [ n^2 - 2n ]`
`= sum n^2 - 2 sum n`
`= (n (n + 1) (2n + 1))/6 - 2 * (n (n + 1))/2`
`= (n (n + 1) [ (2n + 1) - 6 ])/6`
`= (n (n + 1) (2n - 5))/6`
`=` R.H.S. (Proved)
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Problem : 21 / 23
[ Geometric Progression ]
Enter your problem
21.
For geometric progression, find `1 + 1/sqrt(2) + 1/2 + 1/(2*sqrt(2)) + ... 10` terms ( For geometric progression, find `1 + 1/sqrt(x) + 1/x + 1/(x*sqrt(x)) + ... n` terms where x = 2 and n = 10 . )
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| Solution:
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Here `a = 1, r = 1 / sqrt(2), n = 10`
We know that, `S_n = a * (1 - r^n)/(1 - r)`
`:. S_10 = 1 × (1 - (1 / sqrt(2))^10) / (1 - 1 / sqrt(2))`
`= 1 × (1 - (1 / 2)^5) / ((sqrt(2) - 1) / sqrt(2))`
`= sqrt(2) × 1 × (1 - 1 / 32) / (sqrt(2) - 1) `
`= sqrt(2) × 1 × ((32 - 1) / 32) / (sqrt(2) - 1) × (sqrt(2) + 1) / (sqrt(2) + 1)`
`= sqrt(2) × 1 × (31/32) / (2 - 1) × (sqrt(2) + 1)`
`= (31/32) × sqrt(2) × (sqrt(2) + 1)`
`= (31/32) × (2 + sqrt(2))`
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Problem : 22 / 23
[ Geometric Progression ]
Enter your problem
22.
Find `1^2 + 2^2 + ...+ 10^2` , ( Find `a^2 + b^2 + ... + n^2` , where a = 1 , b = 2 and n = 10 . )
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| Solution:
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Here Find ` 1^2 + 2^2 + 3^2 + ... + 10^2 `
`= sum i^2` (where `i = 1,...,10`)
`= (n (n + 1)(2n + 1)) / 6` (where `n = 10`)
`= (10 (10 + 1) (2×10 + 1)) / 6 `
`= (10 × 11 × 21) / 6`
`= 5 × 11 × 7`
`= 385`
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Problem : 23 / 23
[ Geometric Progression ]
Enter your problem
23.
Find `1^2 + 2^2 + ... 10` terms, ( Find `a^2 + b^2 + ... n` terms, where a = 1 , b = 2 and n = 10 . )
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| Solution:
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Here Find ` 1^2 + 2^2 + 3^2 + ... 10 "terms" `
`= sum i^2` (where `i = 1,...,10`)
`= (n (n + 1)(2n + 1)) / 6` (where `n = 10`)
`= (10 (10 + 1) (2×10 + 1)) / 6 `
`= (10 × 11 × 21) / 6`
`= 5 × 11 × 7`
`= 385`
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