Question

in an AP the sum of first 3 terms is 21 and and that of sum of last three terms is 276 then find the sum of first 20 terms

Answer 1

Answer:

990

Step-by-step explanation:

a+(a+d)+(a+2d)= 21

3a+3d= 21 ----------1

(a+17d)+(a+18d)+(a+19d)= 276

3a+54d = 276--------2

By solving equation 1 and 2,

3a+ 3d = 21

3a+54d = 276

___________

-51d = -255

d = 255/51

d = 5.

Put d=5 in equation 1,

3a+3d=21

3a+3(5)=21

3a=21-15

3a=6

a=6/3

a=2.

S20=?

Sn= n/2[2a+(n-1)d]

= 20/2[2(2)+(20-1)5]

=10[4+95]

=10(99)

=990.

Answer 2

• Given, the sum of first and last three terms are 21 and 276 respectively.

       Let a be the first term and d be the common difference of the AP.

• Now,

       a+(a+d)+(a+2d)= 21

       3a+3d= 21 -----------------(1)

       and (a+17d)+(a+18d)+(a+19d)= 276

       3a+54d = 276------------(2)

• By solving equation 1 and 2 by elimination (or any other) method, we get,

        d = 255/51 = 5

        Put d=5 in equation 1,

        3a+3(5)=21

        a=6/3=2

• Now, sum of first 20 terms,

        Sₙ = n/2 {2a+(n-1)d}

        Putting values,

        S= 10 (4+95) = 990.

The answer is 990.