Question

find the sum of the first 19 terms of an ap whose sum is 41 and 13th term 61. Plz answer asap​

Answer 1

AnswerAnswer:

Step-by-step explanation: This is just rough answer but all are correct

Answer 2

The sum of 19 terms of an AP is 931.

Correct Question:

Find the sum of the 19 terms of an AP whose 8th term is 41 and 13th term is 61.

Given:

The 8th term of an AP is 41 and the 13th term is 61.

To Find:

The sum of the first 19 terms of an AP.

Solution:

The nth term of an AP is given by $a_n=a+(n-1)d$, where a is the first term, d is a common difference, and 'n' is the nth term.

Substitute 8 for n and then 41 for $a_8$ into $a_n=a+(n-1)d$.

$a_8=a+(8-1)d\\41=a+7d\,\,\,\,...(1)$

Substitute 13 for n and then 61 for $a_{13}$ into $a_n=a+(n-1)d$.

$a_13=a+(13-1)d\\61=a+12d\,\,\,\,...(2)$

We need to determine a and d from equations (1) and (2). To find them, find a from equation (1).

$a=41-7d$

Substitute the value of 'a' into the equation (2) and find d.

$61=41-7d+12d\\20=5d\\d=4$

Substitute 4 for d into $a=41-7d$ and find a.

$a=41-7\cdt 4\\a=41-28\\=13$

The first term of AP is 13 and the common difference is 4.

The sum of an AP series is given by, $S_n=\frac{n}{2}(2a+(n-1)d)$.

Substitute 13 for a, 4 for d, and 19 for n into the formula of sum and find the sum of the first 19 terms of an AP.

$S_{19}=\frac{19}{2}(2\cdot 13+(19-1)4)\\=\frac{19}{2}(2\cdot 13+(19-1)4)\\=\frac{19}{2}(26+18\cdot 4)\\=\frac{19}{2}(26+72)\\=\frac{19}{2}(98)\\=19\cdot 49\\=931$

Thus, the sum of 19 terms of an AP is 931.