Question

in an ap sum of the first 2 term is 5 and sum of the last 2 term is 129,then find the sequence​

Answer 1

The possible sequences are

$2, 3, ........, 64, 65$

$1.5, 3.5, ........, 63.5, 65.5$

$-13,18,49,80$

$-28.5,33.5,95.5$

Step-by-step explanation:

Let the first term of the AP is a and common difference d

If the no. of the terms are n

Then

The AP will be

$a, a+d, a+2d, .........., a+(n-2)d, a+(n-1)d$

Given, sum of first two terms is 5

Thus,

$a+a+d=5$

$\implies 2a+d=5$   ............. (1)

And, sum of last two terms is 129

Therefore,

$a+(n-2)d+a+(n-1)d=129$

$\implies 2a+2nd-3d=129$

$\implies 5-d+2nd-3d=129$

$\implies 2nd-4d=124$

$\implies nd-2d=62$

$\implies n=\frac{62+2d}{d}$

$\implies n=2+\frac{62}{d}$

n is a natural number > 2, i.e. 62/d should be a whole number

The possible value of n can be

If d  = 1, n = 64

If d = 2, n = 33

If, d = 31, n = 4

if d = 62, n = 3

If d = 1 and n = 64

Then from (1), a = 2

The sequence will be

$2, 3, ........, 64, 65$

If d = 2 and n = 33

Then from (1), a = 1.5

The sequence will be

$1.5, 3.5, ........, 63.5, 65.5$

If d = 31 and n = 4

Then from (1), a = -13

The sequence will be

$-13,18,49,80$

If d = 62 and n = 3

Then from (1), a = -28.5

The sequence will be

$-28.5,33.5,95.5$

Hope this answer is helpful.

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