6. The ${{6}^{th}}$ and ${{17}^{th}}$ terms of an A.P. are $19$ and $41$ respectively, find the ${{40}^{th}}$ term. - Noon Academy

6. The ${{6}^{th}}$ and ${{17}^{th}}$ terms of an A.P. are $19$ and $41$ respectively, find the ${{40}^{th}}$ term.

Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma

6. The ${{6}^{th}}$ and ${{17}^{th}}$ terms of an A.P. are $19$ and $41$ respectively, find the ${{40}^{th}}$ term.

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

Solutions:

Given,

${{a}_{6}}=19$ and ${{a}_{17}}=41$

We know that, ${{n}^{th}}$ term ${{a}_{n}}=a+\left( n-1 \right)d$

So,

${{a}_{6}}=a+\left( 6-1 \right)d$

⇒ $a+5d=19$ …… (i)

Similarity,

${{a}_{17}}=a+\left( 17-1 \right)d$

⇒ $a+16d=41$ …… (ii)

Solving (i) and (ii),

(ii) – (i) ⇒

$a+16d-\left( a+5d \right)=41-19$

$11d=22$

⇒ $d=2$

Using $d$ in (i), we get

$a+5\left( 2 \right)=19$

$a=19-10=9$

Now, the ${{40}^{th}}$ term is given by ${{a}_{40}}=9+\left( 40-1 \right)2=9+78=87$

Therefore the ${{40}^{th}}$ term is $87.$

i

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