Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5: 9. Find the value of m.

Asked by Pragya Singh | 1 year ago |  87

##### Solution :-

Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = $$\dfrac{30}{(m + 1)}$$ ……. (1)

Now,

a1 = a + d

a2 = a + 2d

a3 = a + 3d …

Hence, a7 = a + 7d

am–1 = a + (m – 1) d

According to the question, we have

$$\dfrac{a+7d}{a+(m-1)d}=\dfrac{5}{9}$$

$$\dfrac{1+7\dfrac{30}{(m+1)}}{1+(m-1)\dfrac{30}{m+1}}=\dfrac{5}{9}$$

$$\dfrac{m+1+210}{m+1+30m-30}=\dfrac{5}{9}$$

$$\dfrac{m+211}{31m-29}=\dfrac{5}{9}$$

9m+1899=155m-145

155m - 9m=1899 +145

146m=2044

m = 14

Therefore, the value of m is 14.

Answered by Pragya Singh | 1 year ago

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