Let’s consider a_{1}, a_{2}, … a_{m} be m numbers such that 1, a_{1}, a_{2}, … a_{m}, 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,

a_{1} = a + d

a_{2} = a + 2d

a_{3} = a + 3d …

Hence, a_{7} = a + 7d

a_{m–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 agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).