- Home
- Standard 11
- Mathematics
8. Sequences and Series
easy
If the first, second and last terms of an $A.P.$ be $a,\;b,\;2a$ respectively, then its sum will be
A
$\frac{{ab}}{{b - a}}$
B
$\frac{{ab}}{{2(b - a)}}$
C
$\frac{{3ab}}{{2(b - a)}}$
D
$\frac{{3ab}}{{4(b - a)}}$
Solution
(c) We have first term $A = a$ ……$(i)$
Second term $A + d = b$……$(ii)$
and last term $l = 2a$……$(iii)$
From $(i), (ii)$ and $(iii),$ $ d=(b-a) $ and $n = \frac{b}{{b – a}}$
Then sum $S = \frac{n}{2}[a + l] = \frac{b}{{2(b – a)}}[a + 2a] = \frac{{3ab}}{{2(b – a)}}$
Trick : Let $a = 2,\;b = 3$then the sum $ = 9$ which is given by option $(c).$
Standard 11
Mathematics
