Q. 234.0( 5 Votes )

# In an A.P. the su

Answer :

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 …… 5(2a + 9d) = 80

2a + 9d = 16 ……

Also, sum of next ten terms = 280

S20 – S10 = 280

S20 – 80 = 280 [From 1]

S20 = 360 2a + 19d = 36 ……

On subtracting  from 

2a + 19d – 2a – 9d = 36 – 16

10d = 20

d = 2

Putting d = 2 in 

2a + 18 = 16

a = –1

Hence, AP is

–1, –1 + 2, –1 + 2(2), …

–1, 1, 3, …

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