The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.

Asked by Pragya Singh | 1 year ago |  82

##### Solution :-

Let a1 and d1 be the first term and the common difference of the first arithmetic

progression respectively and a2 and d2 be the first term and the common difference of the second arithmetic progression respectively.

According to the conditions given in the question,

Sum of n terms of first A.P / Sum of n terms of second A.P.

$$\dfrac{5n+4}{9n+6}$$

$$\dfrac{\dfrac{n}{2}[2a_1+(n-1)d_1]}{\dfrac{n}{2}[2a_2+(n-1)d_2]}= \dfrac{5n+4}{9n+6}$$

$$\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}= \dfrac{5n+4}{9n+6}$$

Substitute n = 35  in the equation.

$$\dfrac{2a_1+34d_1}{2a_2+34d_2}= \dfrac{5(35)+4}{9(35)+6}$$

$$\dfrac{a_1+17d_1}{a_2+17d_2}= \dfrac{179}{321}$$

We also know that 18th term of first A.P. /18th term of second A.P.

$$\dfrac{a_1+17d_1}{a_2+17d_2}$$

Then,

18th term of first A.P. /18th term of second A.P.= $$\dfrac{179}{321}$$

Therefore, 179 : 321 is the ratio of 8th term of both the arithmetic progressions.

Answered by Abhisek | 1 year ago

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