Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the first term be a and common difference be d

31st term = a+30d

11th term = a+10d = 38

16th term = a+15d = 73

16th term - 11th term

a+15d - a - 10d = 73-38

5d = 35

d = 7

31st term = (a+15d) + 15d = 73 + 15(7) = 73 + 105 = 178

Answer 2

$\sf \dagger Given :- \dagger$

11th term of AP is 38 and,

16th term of AP is 73.

$\sf \dagger To find :- \dagger$

The 31st term of AP = ?

$\sf \dagger Solution :- \dagger$

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,

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

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

And,

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

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

From eq (i) and eq (ii),

a + 10d = 38 ‿︵‿︵│

⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ |Subtracting

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

-⠀ -⠀ ⠀ -

━━━━━━━━━━━━━━

-5d = -35

$\purple{ \sf⤇ d = \dfrac{-35}{-5} }$

$\orange{ \sf ⤇ d = 7}$

Now,

Substitute the value of d in equation (i),

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

Then,

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

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

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