Consider an A.P. a1,a2,,an; a10. If a2-a1=- 34, an= 14a1, and i=1^n ai= 5252, then i=1^17 ai is equal to

Question

Accepted Answer

Given: The sequence is an A.P. with first term $a_1$ , common difference $d=-\frac{3}{4}$ , last term $a_n=\frac{a_1}{4}$ , and

S_n=\frac{525}{2}

Find: $\sum_{i=1}^{17} a_i$ .

From the last-term formula of an A.P.,

a_n=a_1+(n-1)d

So,

a_1+(n-1)\left(-\frac{3}{4}\right)=\frac{a_1}{4}

which gives

\frac{3a_1}{4}=\frac{3(n-1)}{4}

Hence,

a_1=n-1

Now use the sum formula,

S_n=\frac{n}{2}(a_1+a_n)=\frac{525}{2}

Therefore,

n\left(a_1+\frac{a_1}{4}\right)=525

\frac{5na_1}{4}=525

so

na_1=420

Substituting $a_1=n-1$ ,

n(n-1)=420

Thus,

n=21

and hence

a_1=20

Now,

S_{17}=\frac{17}{2}\left[2a_1+(17-1)d\right]

Substituting $a_1=20$ and $d=-\frac{3}{4}$ ,

S_{17}=\frac{17}{2}\left[40+16\left(-\frac{3}{4}\right)\right]

=\frac{17}{2}\left[40-12\right]

=\frac{17}{2}\times 28

=136

Therefore, $\sum_{i=1}^{17} a_i=136$ , so the correct option is A.

Answer