Question

The sum of first 20 terms of A. p is 400 and the sum of first 40 terms is 1600 find the sum of first 10th term​

Answer 1

Answer:

2,070

Step-by-step explanation:

this answer help you ok

Answer 2

AnSwer:

Let a be the first term and d be the common difference of the given A.P.

★ Sum of the first 20 terms is S(20).

So,

$: \implies \sf \: \: \: S(20) = \dfrac{20}{2}(2a + 19d) \\ \\ : \implies \sf \: \: \:400 = \dfrac{20}{2}( 2a + 19d)\\ \\ : \implies \sf \: \: \: 400 = 10(2a + 19d)\\ \\ : \implies \sf \: \: \: 2a + 19d = 40 ..... (i) \\ \\ : \implies \sf \: \: \: Here, S(40) = \dfrac{40}{2} (2a + 39d) \\ \\ : \implies \sf \: \: \:1600 = 20(2a + 39d)\\ \\ : \implies \sf \: \: \: 2a + 39d = 80 ....(ii)$

Now, further solving these equations,

$: \implies \sf \: \: \:20d = 40 \\ \\ : \implies \sf \: \: \:d = \dfrac{40}{20} \\ \\ : \implies \sf \: \: { \underline{ \boxed{ \sf { \red{\:d = 2 \: }}}}} \: \bigstar$

Now, substitute this value in eqn (2)

$: \implies \sf \: \: 2a + 19d = 40 \\ \\ : \implies \sf \: \:2a + 19(2) = 40 \\ \\ : \implies \sf \: \: 2a + 38 = 40 \\ \\ : \implies \sf \: \: 2a = 40 - 38 \\ \\ : \implies \sf \: \: 2a = 2 \\ \\ : \implies \sf \: \: a = \dfrac{2}{2} \\ \\ : \implies \sf \: { \underline{ \boxed{ \sf { \purple{ \:a = 1 \: }}}}} \: \bigstar$

★ According to the question,

$: \implies \sf \: \: S(10) = [2 \times 1 + (10 - 1)2] \\ \\ : \implies \sf \: \: S(10) = 5[2 + 9 \times 2] \\ \\ : \implies \sf \: \:S(10) = 5[2 + 18] \\ \\ : \implies \sf \: \: S(10) = 5 \times 20 \\ \\ : \implies \sf \: { \underline{ \boxed{ \bf{ \pink {\:S(10) = 100}}}}} \: \bigstar$

Therefore , the sum of its first 10 terms is 100.