Question

Find the 14th term of the A.P whose 11th term is 38 and 16th term is 73

Answer 1

Step-by-step explanation:

p th term = 38 n = 11

first term =( a )

AP = a+(n-1)d

38 =a+(11-1)d

38 =a+10d-d

38 =a+ 10d. eq________( 1 )

p th term 16. n = 16

first term ( a ) = 73

AP = a+( n- 1) d

73 = a+(16-1)d

= a+ ( 15 ) d

= a+ 15d

73 = a + 15d. eq__________( 2 )

73 = a + 15d. eq_________( 2 )

38 = a + 10d. eq_________( 1 )

-___-____-___ subtract the equation ( 1 ) from

35=5d eq ( 2 )

d = 35/5

d = 7

now we have to find ( a ) first term

putting the value of d in eq__ 1 , we get

73 = a + 15d eq_________( 2 )

73 = a + 15 × 7. since ( d = 7 )

a = 73 - 105

a = - 32

now 14 the term = (a + 13d)

( -32+13×7 )

(59)

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