Question

There are two AP’s. The first term of the second AP is 37. The third term of the 1^st AP is 10 more than the 5^th term of the second AP. The 4^th term of the second AP is one more than the 1^st term of the 1^st AP. If the third term of the second AP is 47. Find the third term of the 1^st AP.
1. 67
2. 61
3. 69
4. 56

Answer 1

Correct Answer - Option 1 : 67

Given:

The first term of the second AP is 37. The third term of the 1^st AP is 10 more than the 5^th term of the second AP. The 4^th term of the second AP is one more than the 1^st term of the 1^st AP. The third term of the second AP is 47.

Formula used:

nth term of an AP, T_n = a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term and common difference of the first AP is a1 and d1 respectively.

Let the first term and common difference of the second AP is a2 and d2 respectively.

From the question,

a2 = 37 and

T₃ of first AP – T₅ of second AP = 10

⇒ (a1 + 2d1) – (a2 + 4d2) = 10      ----(1)

T₄ of second AP – T₁ of first AP = 1

⇒ (a2 + 3d2) – a1 = 1      ----(2)

T₃ of second AP = 47 (given)

⇒ a2 + 2d2 = 47

⇒ 37 + 2d2 = 47

⇒ d2 = 5

Putting a2 = 37 and d2 = 5 in equation (2) we get,

(37 + 3 × 5) – a1 = 1

⇒ a1 = 51

Putting all the values of a1, a2 and d2 in equation (1)

(51 + 2d1) – (37 + 4 × 5) = 10

⇒ 2d1 – 6 = 10

⇒ d1 = 8

Third term of first AP, T₃ = a1 + 2d1

⇒ 51 + 2 × 8 = 67

∴ required answer is 67.