Question

find the 31st term of an ap whose 11th term is 38 and 16th term is 73​

Answer 1

Explanation:

➡️Let first term of Ap is a and common difference is d..

➡️by using this formula,

tn=a+(n-1) d

➡️substitute the values,

t11=a+(11-1) d=a+10d=38

➡️same way

t16=a+15d=73

➡️solve this equation then

a=-32 and d=7

➡️now t31=a+30d

=-32+210

=178..

hole it helps..☺❤

Answer 2

$\boxed{\tt \dagger Given :- \dagger}$

11th term of AP is 38 and,

16th term of AP is 73.

$\boxed{ \tt \dagger To find :- \dagger}$

The 31st term of AP = ?

$\boxed{ \tt \dagger Solution :- \dagger}$

Let first term of AP be a

Let first term of AP be aand common difference be d

Let first term of AP be aand common difference be dNow,

$\tt \red{a_{11}=38a}$

$\tt\longrightarrow \green{a+10d=38\:.............(i)}$

And,

$\blue{\tt\:a_{16}=73a}$

$\tt\longrightarrow\pink{a+15d=73\:.............(ii)}$

From eq (i) and eq (ii),

a + 10d = 38 ‿︵‿︵│

⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ |Subtracting

$\boxed{+}$a $\boxed{+}$ 15d = $\boxed{+}$73 ‿︵‿︵│

-⠀ -⠀ ⠀ -

━━━━━━━━━━━━━━

-5d = -35

$\purple{ \tt⤇ d = \dfrac{-35}{-5} }$

$\orange{ \tt⤇ d = 7}$

Now,

Substitute the value of d in equation (i),

$\tt a + 10d = 38 \\ \tt⤇ a + 10 × 7 = 38 \\ \tt⤇ a + 70 = 38 \\ \tt⤇ a = 38 - 70 \\ \tt⤇ a = -32$

Then,

$\gray{\tt\:a_{31}=a+30da }$

$\tt\longrightarrow\:a_{31}=-32+30\times{7} \\ \tt\longrightarrow\:a_{31}=-32+210 \\ \tt\longrightarrow\:a_{31}=178$

Hence, the 31st term of an AP was $\boxed{\sf\pink{178.}}$