The sum of the 2nd and 5th terms of an arithmetic progression (AP) is 42. If the difference between the 6th and 3rd term is 12, find the
(i) Common difference
(ii) first term
(iii) 20th term.
\(T_{n} = a + (n - 1) d\) (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) \(T_{20} = a + 19d\)
= \(11 + 19(4) \)
= \(87\)
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