Question

17) I can't find what this is please help

Answer 1

Heya!

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Now,

First A.P = 2 , 5 , 8 ...

a = 2
d = 3
n = 2n

Sn = 2n / 2 ( 2×2 + (2n - 1) 3 )
Sn = n ( 4 + 6n - 3)
Sn = n ( 1 + 6n )
Sn = n + 6n²

Second A.P = 57 , 59 , 61 ...

a = 57
d = 2
n = n

Sn = n/2 ( 2×57 + (n-1) 2 )
Sn = n/2 (114 + 2n - 2)
Sn = n/2 ( 112 + 2n)
Sn = 56n + n²

Given,

Sn = Sn

Keeping the values

n + 6n² = 56n + n²
n - 56n = n² - 6n²
- 55n = - 5n²
55n = 5n²
55n / 5 = n²
11 n = n²

0 = n² - 11n
0 = n ( n - 11 )

Thus,

n = 0
n = 11

Option ( C ) is correct.


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Hope it helps...!!!

Answer 2

Heya Mate

∆ Warning ! At the end, the Question might yield up two values, one of which would be sane ! Presumably, n = 0 would just be one of the solution herein but note that, ^^" not every human is concerned about the Zeroeth of anything !

So, we shall skip the trivial soln. to meet the options !
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(•) Sum of 'n' terms of an A.P. is given as :

$S_n = \frac{n(2a + (n - 1)d)}{2}$

-> Where d = common difference of the A.P.

-> And a = First term of the A.P.

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(•) Now, according to the Question :-

$\frac{2n(2 \times 2 + (2n - 1)3)}{2} = \frac{n(2 \times 57 + (n - 1)2)}{2}$

$= > n(6n + 1) = n(56 + n)$

$= > 5n(n - 11) = 0$

$= > n = 0 \: or \: n = 11$

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Oyasumi ^^