Question

Find the 31st term of an AP whose 11th term is 38 and 6th term is 73

Answer 1

Answer:

Given Values:

• a₁₁ = 38
• a₆ = 73

To Find:

• a₃₁ = ?

Solution:

⇒ a₁₁ = a + 10d = 38   ...( 1 )

⇒ a₆ = a + 5d = 73    ...( 2 )

Subtracting Equation 1 from Equation 2, we get,

⇒ a + 5d - ( a + 10d ) = 73 - 38

⇒ a + 5d - a - 10d = 35

⇒ -5d = 35

⇒ d = 35 / -5 = -7

Hence, substituting the value of 'd' in Equation 1, we get,

⇒ a + 10 ( -7 ) = 38

⇒ a - 70 = 38

⇒ a = 38 + 70 = 108

Hence a₃₁ = a + 30d

⇒ a₃₁ = 108 + 30 ( -7 )

⇒ a₃₁ = 108 - 210

⇒ a₃₁ = - 102.

Hence the 31st term is - 102.

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