Question

Find the 31st term of a.p whose 11th term is 38 and the 16th term is 73.

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Answer 1

⛄AnSwer :

• The term is 178.

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✒ GiVen :

• a11 = 38
• a16 = 73

✒ SoluTion :

We know that,

$\implies$ an = a + (n – 1)d

$\implies$ a11 = a + 10d = 38

$\implies$ a16 = a + 15d = 73

Now,

Subtracting 11th term from 16th term,

$\implies$ a + 15d – a – 10d

$\implies$ 73 – 38

$\implies$ 5d = 35

$\implies$ d = 7

Substituting the value of d in 11th term we get;

$\implies$ a + 10 x 7 = 38

$\implies$ a + 70 = 38

$\implies$ a = 38 – 70

$\implies$ a = - 32

Now 31st term will be,

$\implies$ a31 = a + 30d

$\implies$ - 32 + 30 x 7

$\implies$ - 32 + 210

$\implies$ 178

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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 ‿︵‿︵│

-⠀ -⠀ ⠀ -

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