Question

16. Find the 31st term of an A.P. whose 10th
term is 38 and 16th term is 74.

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

An A.P. whose 10th term is 38 and 16th term is 74.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The 31th term.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that formula of an A.P.

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

A/q

$\longrightarrow\tt{a_{10}=38}\\\\\\\longrightarrow\tt{38=a+(10-1)d}\\\\\\\longrightarrow\tt{38=a+9d...........................(1)}$

&

$\longrightarrow\tt{a_{16}=74}\\\\\\\longrightarrow\tt{74=a+(16-1)d}\\\\\\\longrightarrow\tt{74=a+15d.........................(2)}$

Subtracting equation (1) & (2),we get;

$\longrightarrow\tt{a+9d-a-15d=38-74}\\\\\\\longrightarrow\tt{9d-15d=-36}\\\\\\\longrightarrow\tt{-6d=-36}\\\\\\\longrightarrow\tt{d=\cancel{\dfrac{-36}{-6} }}\\\\\\\longrightarrow\tt{\green{d=6}}$

Putting the value of d in equation (1),we get;

$\longrightarrow\tt{38=a+9(6)}\\\\\\\longrightarrow\tt{38=a+54}\\\\\\\longrightarrow\tt{a=38-54}\\\\\\\longrightarrow\tt{\green{a=-16}}$

Now;

$\longrightarrow\tt{a_{31}=a+(31-1)d}\\\\\\\longrightarrow\tt{a_{31}=-16+30\times 6}\\\\\\\longrightarrow\tt{a_{31}=-16+180}\\\\\\\longrightarrow\tt{\green{a_{31}=164}}$

Thus;

The 31th term is 164 .

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 $\boxed{\sf\pink{178.}}$