Question

10. The sum of the first and 19th terms of an arithmetic sequence is 60. The sum of its
first and 20th terms is 62.
a) What is its common difference.
b) What is the sum of its 8th and 12th terms.
c) Find its 10th term.​

Answer 1

Answer:-

Given:

Sum of first term and 19th term of an AP = a + a₁₉ = 60

Sum of first term and 20th term = a + a₂₀ = 62

We know that,

nth term of an AP (aₙ) = a + (n - 1)d

So, a + a + (19 - 1)d = 60

⟶ 2a + 18d = 60 -- equation (1)

Similarly,

⟶ 2a + 19d = 62 -- equation (2)

Subtract equation (1) from equation (2).

⟶ 2a + 19d - (2a + 18d) = 62 - 60

⟶ 2a + 19d - 2a - 18d = 2

⟶ d = 2

Substitute the value of d in equation (1)

⟶ 2a + 18(2) = 60

⟶ 2a = 60 - 36

⟶ a = 24/2

⟶ a = 12

Now,

★ a₈ + a₁₂ =

⟶ a + (8 - 1)d + a + (12 - 1)d

⟶ 12 + 7(2) + 12 + 11(2)

⟶ 12 + 14 + 12 + 22

⟶ 60

★ a₁₀ = 12 + (10 - 1)(2)

⟶ 12 + 18

⟶ 30

∴

• common difference is 2.
• Sum of 8th and 12th terms = 60
• 10th term = 30.

Answer 2

Answer:

${ \bold{ \red{Given}}}$

• The sum of the first and 19th terms of an arithmetic sequence is 60
• sum of its first and 20th terms is 62.

${ \bold{ \green{Find}}}$

• What is its common difference.
• What is the sum of its 8th and 12th terms.
• Find its 10th term.

${ \bold{ \pink{Solution}}}$

Sum of first term and 19th term = 60

${ \sf{ \boxed{ a_{n } = a + (n - 1)d} }}$

${ \boxed{ \to{ \sf{ a_{19} = a + 18d}}}}$

${ \implies{a + a + 18d = 60}}$

${ \implies{}2a + 18d = 60.......(1)}$

Sum of first and 20th term = 62

${ \boxed{ \to{ a_{20} = a + 19d}}}$

${ \implies{a + a + 19d = 62}}$

${ \implies{2a + 19d = 62......(2)}}$

Now do equation (2) - (1) :-

${ \implies{2a + 19d - (2a + 18d) = 62 - 60}}$

${ \implies{2a + 19d - 2a - 18d = 2}}$

${ \sf{so \: \: a_{20} - a_{19} = 2}}$

The common difference is 2.

${ \therefore{d = 2}}$

Now substitute the value of d in equation (2)

${ \implies{2a + 19d = 62}}$

${ \implies{2a + 19 \times 2 = 62}}$

${ \implies{2a + 38 = 62}}$

${ \implies{2a = 62 - 38}}$

${ \implies{2a = 24}}$

${ \implies{a = \frac{24}{2} }}$

${ \implies{a = 12}}$

So, the value of a is 12.

(A) 2 is the common difference

(B) sum of 8th and 12th term.

${ \to{ a_{8} + a_{12} }}$

${ \to{a + 7d + a + 11d}}$

${ \to{12 + 7 \times 2 + 12 + 11 \times 2}}$

${ \to{12 + 14 + 12 + 22}}$

${ \to{60}}$

Sum of its 8th and 12th term is 60.

(C) 10th term.

${ \to{ a_{10} = a + 9d}}$

${ \to{ a_{10} = 12 + 9 \times 2 }}$

${ \to{ a_{10} = 12 + 18 }}$

${ \to{ a_{10} = 30}}$

10th term of AP is 30 .