Question

if the sum of a certain number of terms of the A.P.25,22,19,......is 116,find the last term

Answer 1

Here is your answer....

Answer 2

Answer:

Last term in A.P = 4

Step-by-step explanation:

Given A.P :

25,22,19,...

First term (a) = 25

Common difference (d) = a2-a1

= 22-25

= -3

$sum\: of \: n \: terms \\=S_{n}=\frac{n}{2}[2a+(n-1)d]}$

$S_{n}=116$ \* given

$\implies \frac{n}{2}[2\times 25+(n-1)(-3)]=116$

$\implies n(50-3n+3)=232$

$\implies n(53-3n)=232$

$\implies 53n-3n^{2}=232$

$\implies 3n^{2}-53n+232=0$

Splitting the middle term, we get

$\implies 3n^{2}-24n-29n+232=0$

$\implies 3n(n-8)-29(n-8)=0$

$\implies (n-8)(3n-29)=0$

$\implies n-8 =0 \: or \: 3n-29=0$

$\implies n=8 \: or \: n=\frac{29}{3}$

n should not be fraction.

Therefore,

n = 8

Now,

Last term (l) = a+(n-1)d

= 25+(8-1)(-3)

= 25+7(-3)

= 25 -21

= 4

Therefore,

Last term in A.P = 4

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