Question

The 10th
term of an AP is 52 and 16th term is
82 then find the 32nd
term​

Answer 1

Answer:

${ \huge{ \rm{ \red{Solution:}}}}$

From Question,

• 10th term in AP = 52
• a + 9d = 52....(1)

• 16th term in AP = 82
• a + 15d = 82....(2)

By subtracting equation 1 & 2,

$: { \implies{ \sf{a + 9d - (a + 15d) = 52 - 82}}} \\ \\ : { \implies{ \sf{a + 9d - a - 15d = - 30}}} \\ \\ : { \implies{ \sf{{ \cancel{a}} + 9d - \: { \cancel{ a}} - 15d = - 30}}} \\ \\ : { \implies{ \sf{9d - 15d = - 30}}} \\ \\ : { \implies{ \sf{ - 6d = - 30}}} \\ \\ : { \implies{ \sf{ d = \frac{ - 30}{ - 6} }}} \\ \\ : { \implies{ \sf{d = 5}}}$

Now, Substitute d value in equation 1,

$\: : { \implies{ \sf{a + 9d = 52}}} \\ \\ : { \implies{ \sf{a + 9 \times 5 = 52}}} \\ \\ : { \implies{ \sf{a + 45 = 52}}} \\ \\ : { \implies{ \sf{a = 52 - 45}}} \\ \\ : { \implies{ \sf{a = 7}}}$

So, Value of a = 7 , d = 5

Let's Find 32nd term:-

$\:: { \implies{ \sf{ a_{32} = a + 31d }}} \\ \\ : { \implies{ \sf{a_{32} = 7 + 31(5)}}} \\ \\ : { \implies{ \sf{a_{32} = 7 + 155}}} \\ \\ : { \implies{ \sf{a_{32} = 162}}}$

${ \therefore{ \underline{ \pink{ \sf{32nd \: term \: of \: AP \: is \: 162}}}}}$