Question

If in any decreasing arithmetic progression, sum of all its terms, except the first term is equal to 36, the sum of all its terms, except for the last term is zero and the difference of the tenth and the sixth term is equal to - 16, then first term of the series is​

Answer 1

Step-by-step explanation:

Let S be the sum of the series of the A.P. Let 'a' is the first term and 'l' is the last term.

Since it is given that in a decreasing AP the sum of all its terms except the first term is equal to −36, therefore, S=−36+a.

Also since the sum of all its terms except the last term is zero therefore, S=0+l.

and hence 

S=−36+a=0+l

⇒−36+a=0+l

⇒a=l+36

Also it is given that the difference of the tenth and the sixth term is −16. Therefore, T10−T6=−16.

Since, T10=a+(10−1)d=a+9d

T6=a+(6−1)d=a+5d

Substituting the values in T10−T6=−16, we get

T10=a+(10−1)d=a+9d

T6=a+(6−1)d=a+5d

a+9d−a+5d=−16

4d=−16

d=−4

In general, l=a+(n−1)d

Substitute l=a−36 and d=−4 in