The sums of first $n$ terms of two $A . P^{'} s$ are in the ratio $(3 n + 8) : (7 n + 15)$ . The ratio of their $12^{t h}$ terms is

\frac{4}{9}

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\frac{7}{16}

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\frac{3}{7}

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\frac{16}{7}

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Solution

The correct option is B $\frac{7}{16}$
$\begin{matrix} T h e r e a r e 2 A p w i t h d i f f e r e n t f i r s t t e r m a n d c o m m o n d i f f e r e n c e f o r t h e f i r s t A P l e t f i r s t t e r m b e a c o m m o n d i f f e r e n c e b e d s u m o f n t e r m = S_{n} = \frac{n}{2} [2 a + (n - 1) d] n t h t e r m = a_{n} = a + (n - 1) d s i m i a l r l y f o r sec o n d t e r m b e = A c o m m o n d i f f e r e n c e = D S_{n} = \frac{n}{2} [2 A + (n - 1)] D n t h t e r m = A + (n - 1) D \frac{a_{12} o f f i r s t A P}{A_{12} o f 2 n d A P} = \frac{a + (12 - 1) d}{A + (12 - 1) D} = \frac{a + 11 d}{A + 11 D} g i v e n t h a t \frac{s u m o f n t e r m s o f 1 s t A P}{s u m o f n t e r m o f 2 n d A P} = \frac{3 n + 8}{7 n + 15} \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 A + (n - 1) D]} = \frac{3 n + 8}{7 n + 15} \frac{2 (a + (\frac{n - 1}{2}) d)}{2 (A + (\frac{n - 1}{2}) D)} = \frac{3 n + 8}{7 n + 15} \frac{a + (\frac{n - 1}{2}) d}{A + (\frac{n - 1}{2}) D} = \frac{3 n + 8}{7 n + 15} \to (i) w e n e e d t o f i n d \frac{a + 11 d}{A + 11 D} \frac{n - 1}{2} = 11 n - 1 = 22 n = 23 p u t t i n g n i n e q . (i) \frac{a + (\frac{23 - 1}{2}) d}{A + (\frac{23 - 1}{2}) D} = \frac{3 \times 23 + 8}{7 \times 23 + 15} \frac{a + 11 d}{A + 11 D} = \frac{77}{176} \frac{a + 11 d}{A + 11 d} = \frac{7}{16} T h e r a t i o o f 12 t h t e r m i s \frac{7}{16} \end{matrix}$

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12^{t h}

The sums of first n terms of two A.P’s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5^th terms.

The ratio of the sum of n terms of two A.P's is (7n + 1) : (4n + 27). The ratio of their 12^thterms is:

3 n + 8 : 7 n + 15

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