Question 39

The 3rd and 9th term of an arithmetic progression are -8 and 10 respectively. What is the 16th term?

Solution

Let the first term of an AP = $$a$$ and the common difference = $$d$$

3th term of AP = $$A_3=a+2d=-8$$ ----------(i)

9th term = $$A_9=a+8d=10$$ --------(ii)

Subtracting equation (i) from (ii), we get :

=> $$8d-2d=10-(-8)$$

=> $$6d=18$$

=> $$d=\frac{18}{6}=3$$

Substituting it in equation (ii), => $$a=10-8(3)=10-24=-14$$

$$\therefore$$ 16th term = $$A_{16}=a+15d$$

= $$-14+15(3)=-14+45=31$$

=> Ans - (D)

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