Question

In an ap ,the sum of first ten terms is 210 and the sum of next ten terms is 610 ,find the AP
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Answer 1

Answer:

This is the methid. now you put your digits and do by yourself so you also know how to do and plzz mark me as branlist and follow me. Let the first term of AP be ‘a’ and common difference be ‘d’.

We know, sum of ‘n’ terms of an AP is

Given,

Sum of first test ten terms, S10 = 80 ……[1]

⇒ 5(2a + 9d) = 80

⇒ 2a + 9d = 16 ……[2]

Also, sum of next ten terms = 280

⇒ S20 – S10 = 280

⇒ S20 – 80 = 280 [From 1]

⇒ S20 = 360

⇒ 2a + 19d = 36 ……[3]

On subtracting [2] from [3]

2a + 19d – 2a – 9d = 36 – 16

⇒ 10d = 20

⇒ d = 2

Putting d = 2 in [2]

⇒ 2a + 18 = 16

⇒ a = –1

Hence, AP is

–1, –1 + 2, –1 + 2(2), …

–1, 1, 3, …

Answer 2

Answer:

3, 7, 11...

Step-by-step explanation:

Notations used:-

a = First Term

d = Common difference

$S_{10}$ = Sum of 10 terms

$S_{20}$ = Sum of 20 terms

n = number of terms

Sum of first 10 terms = $S_{10}$ = 210

Sum of next 10 terms = 610

So, sum of first 20 terms = 210 + 610 = 820

Sum of n terms = $S_{n}$ = $\frac{n}{2}$ (2a + (n - 1) d)

$S_{10}$ = $\frac{10}{2}$ (2a + (10 - 1) d) = 210

2a + 9d = 210 ×$\frac{2}{10}$

2a + 9d = 21 × 2

2a + 9d = 42 --- (i)

$S_{20}= \frac{20}{2}$ (2a + (20 -1) d) = 820

2a + 19d = 820 × $\frac{2}{20}$

2a + 19d = 82 --- (ii)

Subtracting (ii) from (i),

2a + 19d - (2a + 9d) = 82 - 42

2a + 19d - 2a - 9d = 82 - 42

2a - 2a + 19d - 9d = 82 - 42

10d = 40

d = $\frac{40}{10}$

d = 4

Common difference = d = 4

Substitute d = 4 in (i)

2a + 9d = 42

2a + 9(4) = 42

2a + 36 = 42

2a = 42 - 36

2a = 6

a = $\frac{6}{2}$

a = 3

First term = a = 3

So, the AP is

a = 3

a + d = 3 + 4 = 7

a + 2d = 3 + 4(2) = 3 + 8 = 11

= 3, 7, 11...