Question

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

Answer 1

Answer:

The 31st term of the A.P is 178

Step-by-step explanation:

Let $a_{11}$ = 38

Let $a_{16}$ = 73

Since $a_{n}$ = a + (n-1)d

$a_{11}$ = a + (11-1)d

a + 10d = 38 (eq. 1)

Also, $a_{16}$ = a + (16-1)d

        a + 15d = 73 (eq. 2)

Subtracting eq. 1 from eq. 2

a + 15d = 73

a + 10d = 38

⇒ a - a + 15d - 10d = 73 -38

⇒ 5d = 35

⇒ d = $\frac{35}{5}$

⇒ d = 7

Substituting value of 'd' in eq. 1

a + 15 × (7) = 73

a + 105 = 73

a = 73 - 105

a = -32

So, $a_{31}$ = a + (31-1)d

           = (-32) + 30 × (7)

           = -32 + 210

           = 178

Answer 2

$\boxed{\tt \dagger <strong>Given :-</strong> \dagger}$

11th term of AP is 38 and,

16th term of AP is 73.

$\boxed{ \tt \dagger <strong>To find :-</strong> \dagger}$

The 31st term of AP = ?

$\boxed{ \tt \dagger <strong>Solution :-</strong> \dagger}$

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