Question

Find the sum of first 20 terms of the sequence whose nth term is $a_{n}$ = An + B.

Answer 1

Answer:

The sum of first 20 terms of the given  sequence is 210A + 20B

Step-by-step explanation:

Given :  

nth term , an= An + B………….(1)

On putting n = 1 in eq 1,

a1 = A + B

On putting n = 20 in eq 1,

a20 = A × 20 + B

a20 = 20A + B

a20 = l(last term)  = 20A + B

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

S20 = 20/2 [a + l]

S20 = 10 [A + B + 20A + B]

S20 = 10 [21A + 2B]

S20 = 210A + 20B

Hence, the sum of first 20 terms of the given  sequence is 210A + 20B

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Answer 2

Answer:

$\boxed{ S_{20} = 210A+20B}$

Explanation:

Given

nth term = $a_{n}$ = An + B

To Find:

Sum of first 20 terms of A.P.

Solution

The nth term is,

$a_{n}$ = 2 - 3n.

First term

$a_{1}$ = A(1)+B

$a_{1}$ = A+B

second term

$a_{2}$ = A(2)+B

$a_{2}$ = 2A+B

Third term

$a_{3}$ = A(3)+B

$a_{3}$ = 3A+B

The series of A.P. is as follows

A+B, 2A+B, 3A+B......

We have

a = A+B

d = 2A+B-A-B = A

n = 20

By the identity

$\boxed{ S_{n}= \dfrac{n}{2}[2a+(n-1)d] }$

The sum of 20 terms is

$S_{20} = \dfrac{20}{2}[2(A+B)+(20-1)(A)]$

$S_{20} = 10 [2A+2B+(19)(A)]$

$S_{20} = 10(21A+2B)$

$\boxed{ S_{20} = 210A+20B}$