Question

The 16th term of an ap is 5 times its 3rd term. If its 10th term is 41 then find the sum of its first 15 terms. use elimination meathod

Answer 1

Hope you find it helpful.

Answer 2

$\large{\underline{\rm{\blue{\bf{Given:-}}}}}$

16th term of an A.P. is five times it's third term.

The tenth term is 41

$\large{\underline{\rm{\blue{\bf{To \: Find:-}}}}}$

Find the sum of the first 15 terms if it's 10th term is 41

$\large{\underline{\rm{\blue{\bf{Solution:-}}}}}$

Let us consider 'a' be the first term and d be the common difference of the AP. Then,

$\sf a_{16}=5 \times a_3$

According to the question,

$\implies \sf a+15d=5(a+2d)$

$\implies \sf [a_n=a+(n-1)d]$

$\implies \sf a+15d=5a+10d$

$\implies \sf 4a=5d$

And also,

$\sf a_{10}=41$

$\implies \sf a+9d-41 \qquad ...(2)$

Now, solving equation (1) and (2), we get

$\implies \sf a+9 \times \dfrac{4a}{5} =41$

$\implies \sf \dfrac{5a+36a}{5}=41$

$\implies \sf \dfrac{41a}{5}=41$

$\implies \sf a=5$

By putting a = 1 in equation (1) we get,

$\sf 5d=4 \times 5=20$

$\implies \sf d=4$

Using the formula, we get

$\sf S_n=\dfrac{15}{2} \bigg[2 \times 5 +(15-1) \times 4 \bigg]$

$\implies \sf \dfrac{15}{2}\times (10+56)$

$\implies \sf \dfrac{15}{2} \times 66$

$\implies \sf 495$

Therefore, the sum of the first 15 term is 495