Math, asked by Anku9110, 1 year ago

If the sums of the first 8 and 19 terms of an AP are 64 and 361 respectively, find the common difference and the sum of n terms of the series.

Answers

Answered by mihirsthacker
6

Here is your answer:

ATQ.

S8 = 64 = 4(2a+7d)

64/4 = 2a+7d

16 = 2a+7d------------------(1)

S19 = 361 = 19/2(2a+18d)

361*2/19 = 2a+18d

38 = 2(a+9d)

19 = a+9d = A10 --------------------(2)

From 2

a = 19-9d

Putting value of a in eq-(1)

16 = 2(19-9d) + 7d

16 = 38 - 18d + 7d

-22 = -11d

d = 22/11

d = 2

Therefore,

19 = a+9*2

a = 19-18

a = 1

Sum of n terms  = Sn

= n/2(2*1+(n-1)2)

= n²

MARK IT AS THE BRAINLIEST PLZZZZ..............

Answered by Anonymous
0

Answer:

• Sum of n terms in AP :

Sn = (n/2)[2a + (n- 1)d]

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⇒ S₈ = 64

⇒ 8/2 × (2a + 7d) = 64

⇒ 4 × (2a + 7d) = 64

⇒ 2a + 7d = 16 — eq. ( I )

⇒ S₁₉ = 361

⇒ 19/2 × (2a + 18d) = 361

⇒ 19 × (a + 9d) = 361

⇒ a + 9d = 19 — eq. ( II )

Multiplying eq.( II ) by 2 & Subtracting from eq.( I ) from eq.( II ) :

↠ 2a + 18d - 2a - 7d = 38 - 16

↠ 11d = 22

↠ d = 2

• Substitute d value in eq. ( II ) :

⇒ a + 18 = 19

⇒ a = 19 - 18

⇒ a = 1

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Sum of nth terms of the AP :

↠ Sn = n/2 [2a + (n - 1)d]

↠ Sn = n/2 × [2 × 1 + (n - 1) × 2]

↠ Sn = n/2 × [2 + 2n - 2]

↠ Sn = n/2 × 2n

↠ Sn = n × n

↠ Sn = n²

Sum of nth terms of the AP is n².

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