(i) Given a₁₂ = 37, d = 3, find a and S₁₂.
(i) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Solutions:

(i) Provided that, a₁₂ = 37, d = 3

The formula for the nth term in an AP, as we know, is

a_n = a+(n −1)d,

Putting the given values, as a result we get,

⇒ a₁₂ = a+(12−1)3

⇒ 37 = a+33

⇒ a = 4

Now, the sum of nth term,

S_n = n/2 (a+a_n)

S_n = 12/2 (4+37)

= 246

(ii) Provided that, a₃ = 15, S₁₀ = 125

The formula for the nth term in an AP, as we know, is,

a_n = a +(n−1)d,

Putting the given values, as a result we get,

a₃ = a+(3−1)d

15 = a+2d ………………………….. (i)

Sum of the nth term,

S_n = n/2 [2a+(n-1)d]

S₁₀ = 10/2 [2a+(10-1)d]

125 = 5(2a+9d)

25 = 2a+9d ……………………….. (ii)

Multiplying the eq. (i) by (ii), we will get;

30 = 2a+4d ………………………………. (iii)

Subtracting the eq. (iii) from (ii), we get,

−5 = 5d

d = −1

From equation (i),

15 = a+2(−1)

15 = a−2

a = 17 = First term

a₁₀ = a+(10−1)d

a₁₀ = 17+(9)(−1)

a₁₀ = 17−9 = 8