MCQ
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$\frac{{ab}}{{2(b - a)}}$
- ✓$\frac{{3ab}}{{2(b - a)}}$
- D$\frac{{3ab}}{{4(b - a)}}$
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 $(c).$
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