The sum of the 1st and 2nd terms of an A.P. is 4 and the 10th term is 19. Find the sum of the 5th and 6th terms.
\(T_{n} = a + (n - 1) d\) (terms of an A.P)
\(T_{1} = a\)
\(T_{2} = a + d\)
\(T_{10} = a + 9d\)
\(a + a + d = 2a + d = 4 ... (i)\)
\(a + 9d = 19 ... (ii)\)
(ii) x 2: \(2a + 18d = 38 ... (iii)\)
(iii) - (i) : \(17d = 34 \implies d = 2\)
\(2a + 2 = 4 \implies 2a = 2\)
\(a = 1\)
\(T_{5} + T_{6}\)
= \((a + 4d) + (a + 5d)\)
= \(2a + 9d\)
= \(2(1) + 9(2)\)
= 20
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}