Question

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Answer 1

$\underline{First \ Question} \\ \\ \\ T_3 = 12 \\ \\ T_7 = 36 \\ \\ \\ T_5 = \frac{T_3 + T_7}{2} \\ \\ = \frac{12 + 36}{2} = \frac{48}{2} = 24 \\ \\ \\ \therefore T_5 = 24 \\ \\ \\ Option \ [2] \ is \ the \ answer.$

$\underline{Second \ Question} \\ \\ \\ d = 15 - 5 = 10 \\ \\ \\ \frac{100}{10} = 10 \\ \\ \\ 16 + 10 = 26 \\ \\ \\ \therefore T_{26} \ is \ 100 \ more \ than \ T_{16}. \\ \\ \\ Option \ [4] \ is \ the \ answer.$ ​

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Answer 2

Solution 1.)

T3 = 12

=> a + 2d = 12 ------(1)

T7 = 36

=> a + 6d = 36 -----(2)

On subtracting equation 1 from 2, we get

4d = 24

=> d = 6

Now,

a + 2(6) = 12

=> a +12 = 12

=> a = 0

T5 = a + 4d

= 0 + 4(6)

= 0 + 24

= 24

___________

Solution 2.)

a = 5

d = 15 - 5 = 10

L = T16 + 100

=> a +(n-1)d = a + 15d + 100

=> (n-1)(10) = 15 (10) +100

=> (n-1)(10) = 150 +100

=> (n-1) (10) = 250

=> n - 1 = 25

=> n = 26

So,

26th term of the given AP is 100 more than 16th term