The sum of the 2nd and 5th terms of an arithmetic progression (AP) is 42.

Explanation

\(T_{n} = a + (n - 1) d\) (for an arithmetic progression).

Given: \(T_{2} + T_{5} = 42 \)

\(T_{6} - T_{3} = 12 \)

\(T_{2} = a + d ; T_{5} = a + 4d\)

\(\implies a + d + a + 4d = 42 ; 2a + 5d = 42 ..... (1)\)

(i) \(T_{6} = a + 5d ; T_{3} = a + 2d\)

\(\implies a + 5d - a - 2d = 12 ; 3d = 12\)

\(3d = 12 \implies d = 4\)

(ii) \(2a + 5d = 42 \implies 2a + 5(4) = 42\)

\(2a + 20 = 42 \implies 2a = 22\)

\(a = 11\)

(iii) \(T_{20} = a + 19d\)

= \(11 + 19(4) \)

= \(87\)