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+1 vote
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in Arithmetic Progression by (22.8k points)
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The ratio between the sum of n terms of two A.P.'s is (7n + 1) : (4n + 27). The ratio of their 11th terms is

(a) 125 : 106 

(b) 148 : 111 

(c) 131 : 89 

(d) 127 : 108

1 Answer

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Best answer

Answer : (b) 148 : 111

Let the two A.P.’s be a, a + d, a + 2d, ....... and A, A + D, A + 2D, ....... 

Given ,

\(\frac{n/2[2a + (n-1)d]}{n/2 [2A + (n-1) D]} = \frac{7n+1}{4n+27}\) 

⇒ \(\frac{2a + (n-1)d}{2A + (n-1) D} = \frac{7n+1}{4n+27}\)  ... (i)

Now, we have to find the ratio \(\frac{t_{11}}{T_{11}}\) = \(\frac{a+10d}{A+10D}\) = \(\frac{2a+20d}{2A+20D}\)

Putting n = 21 in (i), we get

\(\frac{2a+20d}{2A+20D}\) = \(\frac{7 \times 21 +1}{4 \times 21 +27}\) = \(\frac{147+1}{84+27} = \frac{148}{111} \) 

t11 : T11 = 148 : 111.

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