Question

if the sum of the first 2n terms of the AP 2,5,8 .....is equal to the sum of the first n terms of the AP 57,59,61....then n equals

Answer 1

Answer:

11

Step-by-step explanation:

AP : 2,5,8 .....

First Term = a =2

Common Difference = $a_2-a_1 =5-2 =3$

Formula of sum of first n terms= $S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 2n

So,  $S_{2n}=\frac{2n}{2}(2a+(2n-1)d)$

$S_{2n}=n(2(2)+(2n-1)3)$

$S_{2n}=n(4+(2n-1)3)$

AP 57,59,61...

First Term = 57

Common Difference = $a_2-a_1 =59-57 =2$

Formula of sum of first n terms= $S_n=\frac{n}{2}(2a+(n-1)d)$

So, $S_n=\frac{n}{2}(2(57)+(n-1)2)$

Now we are given that  the sum of the first 2n terms of the AP 2,5,8 .....is equal to the sum of the first n terms of the AP 57,59,61....

So,  $(4+(2n-1)3)=\frac{1}{2}(2(57)+(n-1)2)$

$2(4+(2n-1)3)=(2(57)+(n-1)2)$

$8+12n-6=114+2n-2$

$12n+2=112+2n$

$12n-2n=112-2$

$10n=110$

$n=11$

Hence the value of n is 11.

Answer 2

Answer:

$n = 11$

Step-by-step explanation:

$i) Given \: first \:A.P:\\2,5,8,...$

$First\:term (a)=2,\\common\: difference (d)=a_{2}-a_{1}\\=5-2=3$

$We\:know \: that :\\\boxed {Sum\:of \:n\:terms (S_{n})=\frac{n}{2}[2a+(n-1)d]}$

$Here,\\Sum\:of\:2n\:terms S_{2n}=\frac{2n}{2}[2\times 2+(2n-1)3]\\=n(4+6n-3)\\=n(1+6n)--(1)$

$ii) Given\:Second \:A.P:\\57,59,61,...,$

$a=57,\\d=59-57=2$

$Sum\:of\:n\:terms\:S_{n}\\=\frac{n}{2}[2\times 57+(n-1)2]\\=\frac{2n}{2}[57+n-1]\\=n(56+n)---(2)$

According to the problem given,

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

$\implies 1+6n=56+n$

$\implies 6n-n=56-1$

$\implies 5n=55$

$\implies n =\frac{55}{5}$

$\implies n = 11$

Therefore,

$n = 11$

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