Question

if the sum of certain numbers of terms of an A. P 25 22 19 .......is116 find the last term
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Answer 1

Step-by-step explanation:

we have,

A.P. is

25,22,19......

S

n

=116

a=25

d=T

2

−T

1

d=22−25

d=−3

Then, we know that

S

n

=

2

n

(2a+(n−1)d)

⇒116=

2

n

(2×25+(n−1)×(−3))

⇒116=

2

n

(50−3n+3)

⇒116=

2

n

(53−3n)

⇒232=n(53−3n)

⇒3n

2

−53n+232=0

⇒3n

2

−(29+24)n+232=0

⇒3n

2

−29n−24n+232=0

⇒n(3n−29)−8(3n−29)=0

⇒(3n−29)(n−8)=0

If

3n−8=0

n=

3

8

(notpossible)

If

n−8=0

n=8

So,

The last term is

S

n

=

2

n

(a+l)

116=

2

8

(25+l)

116=4(25+l)

116=100+4l

116−100=4l

4l=16

l=4

Hence, the last term of this series is

4.

This is the answer.

.

.

.

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Answer 2

Answer:

a= 25

d = 25-22 = -3

Sn = 116

n/2 [2a + (n-1)d] = 116

n/2 [50 - 3n +3] = 116

n[53 - 3n]= 232

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

therefore, n = 58/6 or 8

hence n= 8

$a_{n}$ = $a_{8}$ = a+ 7d = 25 + 8(-3)= 25-24 = 1

Last term is 1