Question

First term of the arithmetic sequence is 40 and sum of first 6 terms is 290. a) find 3rd term?

Answer 1

Answer:

$\large\boxed{\star\:\:\:\:\:\:a_3=60\:\:\:\:\:\:\star}$

Step-by-step explanation:

✧Given:-

a(first term)=40, S₆(sum of first 6 terms)=290.

✧To Find:-

a₃(3rd term) of its A.P.

✧Formula Applied:-

• aₙ=a+(n-1)d

• Sₙ=$\frac{n}{2}$[2a+(n-1)d]

✧Solution:-

   Sₙ=$\frac{n}{2}$[2a+(n-1)d] ,where S₆=290, n=6, a=40.

⇒290=$\frac{6}{2}$×[2(40)+(6-1)d]

⇒290=3×(80+5d)

⇒290=240+15d

⇒5d=290-240

⇒5d=50

⇒d=$\frac{50}{5}$

⇒d=10

Now, we have to find a₃,

⇒aₙ=a+(n-1)d, where a=40, n=3, d=10.

⇒a₃=a+(3-1)d

⇒a₃=40+(2×10)

⇒a₃=40+20

$\implies\boxed{a_3=60}$