Question

The sums of n terms of 2A.P.'s are in the ratio 7n+1:4n+27. Show that ratio of their 11th term is 4:3.​

Answer 1

Answer:-

Given:-

Ratio of sum of first n terms of two APs is 7n + 1 : 4n + 27.

Let,

a₁ and d₁ be the first term and common difference of the first AP. And, a₂ and d₂ be the first term and common difference of second AP.

We know that,

Sum of first n terms of an AP = n/2 [ 2a + (n - 1)d ]

So, According to the question;

$\implies \sf \: \dfrac{ \cancel{ \frac{n}{2}}(2a_1 + (n - 1)d_1) }{ \cancel{\frac{n}{2}}(2a_2 + (n - 1)d_2) } = \frac{7n + 1}{4n + 27}$

Now, divide both numerator and denominator by 2.

$\: \implies \sf \: \dfrac{ \frac{2a_1 + (n - 1)d_1}{2} }{ \frac{2a_2 + (n - 1)d_2}{2} } = \frac{7n + 1}{4n + 27} \\ \\ \\ \implies \sf \: \frac{a_1 + \big( \frac{n - 1}{2} \big)d_1 }{a_2 + \big(\frac{n - 1}{2} \big)d_2} = \frac{7n + 1}{4n + 27}$

Now,

We have to prove that:-

a₁₁ / a₁₁' = 4/3.

We know that,

nth term of an AP = a + (n - 1)d

So, 11th term = a + (11 - 1)d = a + 10d.

From the above condition;

⟹ (n - 1)/2 = 10

⟹ n - 1 = 20

⟹ n = 20 + 1

⟹ n = 21

Now, substitute n = 21 to find the ratio of their 11th terms.

$\: \implies \sf \dfrac{a_1 + \big( \frac{21 - 1}{2} \big)d_1 }{a_2 + \big(\frac{21 - 1}{2} \big)d_2} = \frac{7(21)+ 1}{4(21) + 27} \\ \\ \\ \implies \sf \frac{a_1 + \big( \frac{20}{2} \big)d_1 }{a_2 + \big(\frac{20}{2} \big)d_2} = \frac{147+ 1}{84+ 27} \\ \\ \\ \implies \sf \frac{a_1 + 10d_1 }{a_2 + 10d_2} = \frac{148}{111}\\ \\ \\ \implies \sf \: \frac{a11}{a11 ^{'} } = \frac{4}{3}$

Hence, Proved.

Answer 2

Solution :-

We know that

Sₙ = n/2[2a + (n - 1)d]

$\sf \dfrac{S_n}{S_n`} = \dfrac{n/2[2a + (n - d)]}{n/2[2a + (n - d)]}$

$\sf \dfrac{7n + 1}{4n + 27} = \dfrac{n/2[2a + (n - 1)d]}{n/2[2a + (n - 1)d]}$

By taking 2 common is nth term and a

$\sf \dfrac{7n + 1}{4n + 27} = \dfrac{[2a + (n - 1)d]/2}{[2a + (n - 1)d]/2}$

$\sf \dfrac{7n + 1}{4n + 27} = \dfrac{[2a + (n - 1)/2]d}{[2a + (n - 1)/2]d}$

$\sf\dfrac{7n+1}{4n+27} = \dfrac{[a + (n-1)/2]d}{[a'+ (n - 1)/2]d}$

Now

Cancelling  n - 1

According to the question

$\sf\dfrac{n - 1}{2} = 10$

$\sf n - 1 = 2 \times10$

$\sf n-1=20$

$\sf n =20+1$

$\sf n =21$

Now

By putting the value

$\sf\dfrac{7n+1}{4n+27} = \dfrac{[a + (21-1)/2]d}{[a'+ (21 - 1)/2]d}$

$\sf\dfrac{7n+1}{4n+27} = \dfrac{[a + (20)/2]d}{[a'+ (20)/2]d}$

$\sf\dfrac{7n+1}{4n+27} = \dfrac{a + 10d}{a'+ 10d'}$

$\sf \dfrac{T11}{T'11} = \dfrac{a + 10 d}{a + 10d}$

$\sf \dfrac{4}{3} = \dfrac{7(21)+1}{4(21)+27}$

$\sf \dfrac{4}{3} = \dfrac{147+1}{84+27}$

$\sf \dfrac{4}{3} = \dfrac{148}{111}$

$\sf\dfrac{4}{3} = \dfrac{4}3$