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
(a) 15
(b) 14 (c) 16 (d) 17.
Answers
Answer:
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, T
10
−T
6
=−16.
Since, T
10
=a+(10−1)d=a+9d
T
6
=a+(6−1)d=a+5d
Substituting the values in T
10
−T
6
=−16, we get
T
10
=a+(10−1)d=a+9d
T
6
=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 l=a+(n−1)d
36=(n−1)(4)
n−1=9
n=10
So, n=10
We got d=−4 and n=10 and it is mentioned as decreasing AP.
From the given sum which was 0 except the last term and negative except the first term, so 'a' has to be positive.
Let a=16 then the AP becomes
16,12,8,4,0,−4,−8,−12,−16,−20 which satisfies all the above conditions and therefore the first term of the AP is 16.