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Home / Problem Set 3

In an A.P. the first term is –5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?

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Sol. For the given A. P.
First term a = –5

t_n = 45
S_n = 120
∴ t_n = a + (n –1)d
45 = –5 + (n –1)d
45 + 5 = (n –1)d
(n –1)d = 50 ...(I)

S_n = \({n\over2}[2a+(n-1)d]\) [Formula]

120 = \({n\over2}[2*(-5)+50]\) [From (I)]

240 = n [ –10 + 50]
240 = n × 40
n = 6
Place n = 6 in eqn. (I)
(6 – 1)d = 50

d = \(50\over 5\)

d = 10

∴ There are 6 terms in the A. P. and the common difference is 10.

Arithmetic Progression September 28 , 2018 0 Comments 18 views

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In an A.P. the first term is –5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?

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