Question

In an A.P., if the 12th term is -13 and the sum of its first four terms is 24, Find the sum of its first ten terms.

Answer 1

`a_(12)=-13`

`a_1+a_2+a_3+a_4=24`

Since the progression is arithmetic, we have

`a_2=a_1+d`

`a_3=a_2+d=a_1+2d`

`a_4=a_3+d=a_1+3d`

So we have

`a_1+(a_1+d)+(a_1+2d)+(a_1+3d)=4a_1+6d=24\implies2a_1+3d=12`

Continuing the pattern above, we also find that

`a_(12)=a_(11)+d=\cdots=a_1+11d\implies a_1+11d=-13`

Writing
`a_1=-11d-13` and substituting into the first equation, we have

`2(-11d-13)+3d=12\implies-19d-26=12\implies19d=-38\implies d=-2`

`\implies a_1-22=-13\implies a_1=9`

Now the sum of the first ten terms is

`a_1+a_2+\cdots+a_(10)=10a_1+(0+1+2+\cdots+9)d`

`=10\cdot9+\frac{9\cdot10}2\cdot(-2)=0`