Question

If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.​

Answer 1

To Find :-

The sum of first 20 terms of the AP.

Given :-

⠀⠀⠀⠀$\bigstar$ Third term of the AP.

⠀⠀⠀⠀$\bigstar$ Ratio of 6th term and 10th ⠀⠀⠀⠀⠀⠀term = 7 : 13.

We know :-

⠀⠀⠀⠀⠀⠀n term of an AP :-

$\underline{\bf{a_{n} = a_{1} + (n - 1)d}}$

Where :-

• $a_{n}$ = n term of the AP.
• $a_{1}$ = First term of the AP.
• $d$ = Common Difference.

⠀⠀⠀⠀⠀⠀⠀Sum of nth term :-

$\underline{\boxed{\bf{s_{n} = \dfrac{n}{2}\bigg[2a_{1} + (n - 1)d\bigg]}}}$

Where :-

• $s_{n}$ = sum of nth term of the AP.
• $a_{1}$ = First term of the AP.
• $d$ = Common Difference.

Concept :-

According to the Question , we have to find the sum of 20 terms of the AP , so first we need to find the common difference and the first term of the AP.

A/c, the sum of 3rd term and the ratio of 6th term to 10th term is given, so we can formed two Equations and by linearly solving them , we will get the required value.

Solution :-

⠀⠀⠀⠀⠀⠀⠀Equation (i) :

From the formula for n term of the AP , we get the 3rd term of the AP as :

$:\implies \bf{a_{n} = a_{1} + (n - 1)d} \\ \\ \\ :\implies \bf{a_{3} = a_{1} + (3 - 1)d} \\ \\ \\ :\implies \bf{a_{3} = a_{1} + 2d}$

Thus, the 3rd term of the AP is $\bf{a_{3} = a_{1} + 2d}$

We also know that the 3rd term of the AP is 5 , so we get the Equation as :-

$:\implies \bf{5 = a_{1} + 2d}$ [Eq.(I)]

Hence, the Equation formed is $\bf{5 = a_{1} + 2d}$

⠀⠀⠀⠀⠀⠀⠀⠀Equation (ii) :

From the formula for n term of the AP , we get the 5th and 10 th term of the AP as :

• 5th term of the AP

$:\implies \bf{a_{n} = a_{1} + (n - 1)d} \\ \\ \\ :\implies \bf{a_{6} = a_{1} + (6 - 1)d} \\ \\ \\ :\implies \bf{a_{6} = a_{1} + 5d}$

Hence, the 5th term of the AP is $\bf{a_{6} = a_{1} + 5d}$

• 10th term of the AP

$:\implies \bf{a_{n} = a_{1} + (n - 1)d} \\ \\ \\ :\implies \bf{a_{10} = a_{1} + (10 - 1)d} \\ \\ \\ :\implies \bf{a_{10} = a_{1} + 9d}$

Hence, the 10th term of the AP is $\bf{a_{10} = a_{1} + 9d}$

According to the Question, the ratio of 6th to 10th term is 7 : 13 , so we get :

$\bf{\dfrac{a_{1} + 5d}{a_{1} + 9d} = \dfrac{7}{13}}$

On solving the above equation , we get :-

$:\implies \bf{\dfrac{a_{1} + 5d}{a_{1} + 9d} = \dfrac{7}{13}} \\ \\ \\ :\implies \bf{13(a_{1}) + 5d) = 7(a_{1} + 9d)} \\ \\ \\ :\implies \bf{13a_{1} + 65d = 7a_{1} + 63d} \\ \\ \\ :\implies \bf{(13a_{1} - 7a_{1}) + (65d - 63d) = 0} \\ \\ \\ :\implies \bf{6a_{1} + 2d = 0}$

Hence, equation (ii) formed is $\bf{6a_{1} + 2d = 0}$

First term and common difference of the AP :

Now, by putting the two Equations together and solving them , we get :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{a_{1} + 2d = 5}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{6a_{1} + 2d = 0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀___________

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{-5a_{1} = 5}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \bf{a_{1} = \dfrac{5}{-5}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \bf{a_{1} = -1}$

Hence, the first term of the AP is (-1).

Now , putting the value of first term in the equation (i) , we get :-

$:\implies \bf{a_{1} + 2d = 5}$

$:\implies \bf{-1 + 2d = 5}$

$:\implies \bf{2d = 5 + 1}$

$:\implies \bf{2d = 6}$

$:\implies \bf{\not{2}d = \not{6}}$

$:\implies \bf{d = 3}$

Hence, the common difference of the AP is 3.

⠀⠀⠀Sum of 20 terms of the AP :

• $a_{1}$ = (-1)
• $d$ = 3
• n = 20

Using the formula and substituting the values in the equation, we get :

$:\implies \bf{s_{n} = \dfrac{n}{2}\bigg[2a_{1} + (n - 1)d\bigg]} \\ \\ \\ :\implies \bf{s_{20} = \dfrac{20}{2}\bigg[2 \times (-1) + (20 - 1)3\bigg]} \\ \\ \\ :\implies \bf{s_{20} = 10[-2 + 19 \times 3]} \\ \\ \\ :\implies \bf{s_{20} = 10[-2 + 57]} \\ \\ \\ :\implies \bf{s_{20} = 10 \times 55} \\ \\ \\ :\implies \bf{s_{20} = 550} \\ \\ \\ \therefore \bf{s_{20} = 550}$

Hence, the sum of 20 terms of the AP is 550.