Question

Let an be an arithmetic progression with a4 + a7 + a10 = 17 and a4 + a5 +···+ a13 + a14 = 77. If ak = 13, then k =?

Answer 1

Answer:

The value of k is 18.

Step-by-step explanation:

Here, the AP is,

$a_1, a_2,........a_n$

Let a be the first term and d is the common difference,

So the arithmetic sequence would be,

$a, a+d, a+2d, ..........a+(n-1)d$

Given,

$a_4 + a_7 + a_10 = 17$

$\implies a+3d+a+6d+a+9d=17$

$3a+18d=17------(1)$

Now, $a_4 + a_5 +.......+ a_{13}+ a_{14}= 77$

$\implies \frac{11}{2}(2(a+3d)+(11-1)d)=77$

$11(2a+6d+10d)=154$

$22a+176d=154-----(2)$

22 × equation (1) - 3 × equation (2),

We get,

$396d-528d = 374 - 462$

$-132d=-88$

$\implies d=\frac{88}{132}=\frac{2}{3}$

From equation (1),

$3a+\frac{36}{3}=17$

$3a+12=17$

$3a=5$

$a=\frac{5}{3}$

Here,

$a_k=13$

$a+(k-1)d=13$

$(k-1)d=13-a$

$k-1=\frac{13-a}{d}$

$k=\frac{13-a}{d}+1$

By substituting the value,

$k=\frac{13-\frac{5}{3}}{\frac{2}{3}}+1=\frac{39-5}{2}+1=17+1=18$

Hence, the value of k is 18.