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

The value of k is 18.

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 (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=(88)/(132)=(2)/(3)`

From equation (1),

`3a+(36)/(3)=17`

`3a+12=17`

`3a=5`

`a=(5)/(3)`

Here,

`a_k=13`

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

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

`k-1=(13-a)/(d)`

`k=(13-a)/(d)+1`

By substituting the value,

`k=(13-(5)/(3))/((2)/(3))+1=(39-5)/(2)+1=17+1=18`

Hence, the value of k is 18.