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