Question

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

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\large{\underline{\red{\rm{Let:}}}}$

• $\rm{The\: first\:term\:AP=a}$
• $\rm{The\: common\: difference=d}$

$\large{\underline{\red{\rm{Given:}}}}$

• $\rm{The\:10th\:term\:of\:AP=38}$
• $\rm{The\:16th\:term\:of\:AP=74}$

$\large{\underline{\purple{\rm{To\: Find:}}}}$

• $\rm{The\:31th\:term\:of\:AP}$

$\small{\underline{\purple{\rm{As\:per\:the\: above\: information:}}}}$

$\rm\green{\boxed{T_n=a+(n-1)d}}$

$\rm{\implies T_{10}=a+(10-1)d}$

$\rm{\implies 38=a+9d}$

$\rm\purple{\implies a+9d=38------(i)}$

$\rm{\implies T_{16}=a+(16-1)d}$

$\rm{\implies 74=a+15d}$

$\rm\purple{\implies a+15d=74------(ii)}$

• [solving equations here]

$\rm{\implies a+9d=38}$

$\rm{\implies a+15d=74}$

• $\rm{by\: solving\: equation\:we\:get}$

$\rm{\implies -6d=-36}$

$\rm\red{\implies d=6}$

• [putting the value of d=6 in EQ 1

$\rm{\implies a+9d=38}$

$\rm{\implies a+9*6=38}$

$\rm{\implies a+54=38}$

$\rm{\implies a=38-54}$

$\rm\red{\implies a=-16}$

Hence,

$\rm{\implies T_{31}=a+(n-1)d}$

$\rm{\implies T_{31}=-16+(31-1)6}$

$\rm{\implies T_{31}=-16+30*6}$

$\rm{\implies T_{31}=-16+180}$

$\rm\purple{\implies T_{31}=164}$

Answer 2

GIVEN:

The 10th term of an AP = a10 = 38

The 16th term of an AP = a16 = 74

TO FIND:

The 31st term of AP

SOLUTION:

10th term = a + 9d = 38 ---(1)

16th term = a + 15d = 74 --(2)

Solve the both equations

a + 9d - a + 15d = 38 - 74

==> -6d = -36

==> d = 36/6

==> d = 6

Common Difference = 6

Substitute (d) in eq - (1) to find first term (a).

==> a + 9d = 38

==> a + 9(6) = 36

==> a + 54 = 36

==> a = 36 - 54

==> a = -16

First term = -16

We know that,

an = a + (n - 1)d

a31 = -16 + (31 - 1)6

= -16 + (30)6

= -16 + 180

= 164

Therefore, the 31st term of AP is 164.