If the $21^{s t}$ term of an $A P$ is $50.$ Find the sum of the first $41$ terms of the $A P .$

Open in App

Solution

$T_{21} = 50 = a + (21 - 1) d = a + 20 d$

$S_{41} = \frac{41}{2} (2 a + (41 - 1) d) = \frac{41}{2} (2 a + 40 d) = \frac{41}{2} (2 (a + 20 d)) = 41 (a + 20 d) = 41 \times 50 = 2050$

Sum of 41 terms are $2050$

Suggest Corrections

Similar questions

4 m^{2} - m

n^{t h}

107,

n

21^{s t}

A P

A P

A P

2,

20

A P .

In an AP of 50 terms the sun of first 10 terms is 210 and the sum of last 15 terms in 2565.Find the AP

The 16th term of an AP is 5 times its 3rd term. If its 10th term is 41, find the sum of its first 15 terms.

Join BYJU'S Learning Program