Question

if the nth term of an AP is 9-5n . Find the sum of the first 15 terms

Answer 1

Hope it helps you!

An=9-5n

putting n=1,a1=9-5(1) =4

a2 = 9-5(2) =-1

d=a2-a1 = -1-4=-5

S15=15/2{(2(4)+(15-1)-5}

=15/2(8+(-70)

Answer 2

Answer:

Sum of 15 terms in given A.P

Sum of 15 terms in given A.P= -465

Step-by-step explanation:

Given

$n^{th} \: term \: in \: an \: A.P\\= a_{n}=9-5n$----(1)

Now ,

substitute n=1 in equation (1), we get

i) First term(a) = $a_{1}$

= 9-5×1

= 9-5

= 4-----(2)

ii) Substitute n = 15, we get

$a_{15}=9-5×15$

= $9-75$

=$-66$ -----(3)

________________________

Sum of n terms =$S_{n}$

= $\frac{n}{2}[a+a_{n}]$

__________________________

Sum of 15 terms

= $S_{15}$

= $\frac{15}{2}(4-66)$

/* From (1) & (2) */

= $ \frac{15}{2}(-62)$

=$ 15 \times (-31)$

=$-465$

Therefore,

Sum of 15 terms in given A.P

= -465

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