Question

Find the 31" term of an A.P whose 10 term is 40 and 15th term is 60.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Let the first term be a and common difference be d.

According to the question,

a+(10-1)d = 40 and a+(15-1)d = 60

=> a+9d = 40...(i) and a+14d = 60...(ii)

Subtracting (ii) from (i), we get

5d = 20=> d = 4

Substituting the value of d in (i), we get

a=4

So, 31st term = a+(n-1)d = 4+(31-1)4 = 4+120 = 124

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