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15. Find the second term and the {{n}^{th}} term of an A.P. whose {{6}^{th}} term is 12 and the {{8}^{th}} term is 22.
Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma
15. Find the second term and the {{n}^{th}} term of an A.P. whose {{6}^{th}} term is 12 and the {{8}^{th}} term is 22.

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Given, in an A.P

{{a}_{6}}=12 and {{a}_{8}}=22

We know that {{a}_{n}}=a+\left( n-1 \right)d

So,

{{a}_{6}}=a+\left( 6-1 \right)d=a+5d=12 …. \left( i \right)

And,

{{a}_{8}}=a+\left( 8-1 \right)d=a+7d=22 ……. \left( ii \right)

Solving \left( i \right) and \left( ii \right), we have

\left( ii \right) – \left( i \right) ⇒

a+7d-\left( a+5d \right)=22-12

2d=10

d=5

Putting d in \left( i \right) we get,

a+5\left( 5 \right)=12

a=12-25

a=-13

Thus, for the A.P: a=-13 and d=5

So, the {{n}^{th}} term is given by {{a}_{n}}=a+\left( n-1 \right)d

{{a}_{n}}=-13+\left( n-1 \right)5=-13+5n-5

⇒ {{a}_{n}}=5n-18

Hence, the second term is given by {{a}_{2}}=5\left( 2 \right)-18=10-18=-8

i

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