Question

If the ratio of the sum of the first n terms of two A.Ps is (7n+1):(4n+27) , then find the ratio of their m:nth terms.
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Answer 1

Answer:

Required ratio = 24 : 19.

Step-by-step explanation:

Solution

For 1st A.P.,

$\bf first \: term = a_{1} \\ \bf common \: difference =d _1 \\ \bf sum=S _1$

For 2nd A.P.

$\bf \: first \: term= a_{2} , \\ \bf common \: difference = d_{2}, \\ \bf Sum = S_{2}$

Now ,

$\sf \: \frac{S_1}{S_2} = \frac{ \frac{n}{2} \lgroup2a_1 +( n - 1) d_1 \rgroup}{\frac{n}{2} \lgroup2a_2 +( n - 1) d_2 \rgroup} = \frac{ \lgroup2a_1 +( n - 1) d_1 \rgroup}{ \lgroup2a_2 +( n - 1) d_2 \rgroup}$

$\sf \rightarrow \: \frac{\lgroup2a_1 +( n - 1) d_1 \rgroup}{\frac{n}{2} \lgroup2a_2 +( n - 1) d_2\rgroup} = \frac{7n + 1}{4n + 27}$

To find ratio of 9th terms, we will replace n by 17 i.e. (2 × 9 - 1)

$\sf \: \frac{2a_1 + (17 - 1)d_1}{2a_2 + (17 - 1)d_2} = \frac{7 \times 17 + 1}{4 \times 17 + 27}$

$\sf \rightarrow \: \frac{2a_1 + 16d_1}{2a_2 + 16d_2} = \frac{120}{95}$

$\sf \rightarrow \frac{a _1 +8d _1}{a _2 +8d _2} = \frac{24}{19}$

$\bf \red{{Thus, \: ratio \: of \: 9th \: terms = \frac{24}{19} }}$

Hope it helped you. ⚘