Question

Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.​

Answer 1

${\huge {\overbrace {\underbrace{\blue{ANSWER: }}}}}$

Given that,

11th term, a11 = 38

and 16th term, a16 = 73

We know that,

an = a+(n−1)d

a11 = a+(11−1)d

38 = a+10d ………………………………. (i)

In the same way,

a16 = a +(16−1)d

73 = a+15d ………………………………………… (ii)

On subtracting equation (i) from (ii), we get

35 = 5d

d = 7

From equation (i), we can write,

38 = a+10×(7)

38 − 70 = a

a = −32

a31 = a +(31−1) d

= − 32 + 30 (7)

= − 32 + 210

= 178

Hence, 31st term is 178

Answer 2

Given :-

• 11th term of AP is 38 and,

• 16th term of AP is 73.

To find :-

• The 31st term of AP = ?

Solution :-

Let first term of AP be a

and common difference be d

Now,

$\rm\:a_{11}=38$

$\rm\longrightarrow\:a+10d=38\:.............(i)$

And,

$\rm\:a_{16}=73$

$\rm\longrightarrow\:a+15d=73\:.............(ii)$

From eq (i) and eq (ii),

a + 10d = 38 ‿︵‿︵│

⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ |Subtracting

a + 15d = 73 ‿︵‿︵│

-⠀ -⠀ ⠀ -

━━━━━━━━━━━━━━

-5d = -35

⤇ d = $\rm\dfrac{-35}{-5}$

⤇ d = 7

Now,

Putting the value of d in equation (i),

a + 10d = 38

⤇ a + 10 × 7 = 38

⤇ a + 70 = 38

⤇ a = 38 - 70

⤇ a = -32

Then,

$\rm\:a_{31}=a+30d$

$\rm\longrightarrow\:a_{31}=-32+30\times{7}$

$\rm\longrightarrow\:a_{31}=-32+210$

$\rm\longrightarrow\:a_{31}=178$

Hence,the 31st term of an AP will be 178,