Question

5th term of an arithmetic sequence is 39 and the 29th term is 56. which term is 218​

Answer 1

Answer:-

$\sf{A_{15}=39}$

$\sf{A_{29}=56}$

We know that,

$\bf{A_{n}=a + (n - 1)d}$

Hence,

$\leadsto\sf{A_{15}=a + (15 - 1)d}$

$\leadsto\sf{39=a + 14d}$

And,

$\leadsto\sf{A_{29}=a + (29 - 1)d}$

$\leadsto\sf{56=a + 28d}$

Now, let's find a and d by elimination method,

$\sf{39=(a+14d)}$

$\sf{56=(a+28d)}$

_________________________

$\sf{-17=(-14d)}$

$\sf{17=(14d)}$

So, d = 17/14

So, $\sf{39=(a+14\times\dfrac{17}{14})}$

$\sf{a=39-17}$

So, a = 22

$\rule{200}2$

Now, Finding the term which is 218.

$\bf{A_{n}=a + (n - 1)d}$

Hence,

$\leadsto\sf{218=22 + (n - 1)\dfrac{17}{14}}$

$\leadsto\sf{218 - 22 = (n - 1)\dfrac{17}{14}}$

$\leadsto\sf{196= (n - 1)\dfrac{17}{14}}$

$\leadsto\sf{196=\dfrac{17}{14}n - \dfrac{17}{14}}$

$\leadsto\sf{196+\dfrac{17}{14}=\dfrac{17}{14}n}$

$\leadsto\sf{\dfrac{2744+17}{14}=\dfrac{17}{14}n}$

$\leadsto\sf{\dfrac{2761}{14}=\dfrac{17}{14}n}$

$\leadsto\sf{\dfrac{2761}{14}\times\dfrac{14}{17}=n}$

$\leadsto\sf{n=\dfrac{2761}{17}}$

Note:- "n" doesn't belong to natural number. Hence, the term doesn't exist in the sequence.

$\rule{200}2$