Question

find the 31st term of AP whose 11th term is 38 and the 16th term is 73​

Answer 1

Required Answer:-

Given:

• 11th term = 38
• 16th term = 73

By using formula:

$\boxed{ \sf{an = a + (n - 1)d}}$

where,

• a_n = nth term of the AP
• a = first term of the AP
• n = no. of terms
• d = common difference

Putting the values in the formula,

➙ a11 = 38

➙ a + (11 - 1)d = 38

➙ a + 10d = 38 ------- eq. (1)

and,

➙ a16 = 73

➙ a + (16 - 1)d = 73

➙ a + 15d = 73 ------- eq. (2)

Subtracting both the equations,

➙ a + 15d - (a + 10d) = 73 - 38

➙ a + 15d - a - 10d = 35

➙ 5d = 35

➙ d = 7

➙ Then, a = 73 - 105 = -32

To find:

The 31st term of the AP : a31

= a + (31 - 1)d

= a + 30d

= -32 + 30 × 7

= 178

= 178 ( Answer )

Answer 2

Given that,

→ 11th term, a11 = 38

→ 16th term, a16 = 73

Find out:—

We have to find out the 31st term,

We know that,

→ an = a+(n−1)d

→ a11 = a+(11−1)d

→ 38 = a+10d __________________(i)

In the same way,

→ a16 = a +(16−1)d

→ 73 = a+15d __________________(ii)

On subtracting equation (i) from (ii), we get

→ 35 = 5d

→ d = 35/5

→ d = 7

From equation (i), we can write,

→ 38 = a + 10 × (7)

→ 38 − 70 = a

→ a = −32

→ a31 = a +(31−1) d

→ − 32 + 30 (7)

→ − 32 + 210

→ 178 Answer....

Hence,

→ 31st term is 178.

$@vikram$