Question 89

The 4th and 7th term of an arithmetic progression are 11 and -4 respectively. What is the 15th term?

Solution

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

4th term of AP = $$A_4=a+3d=11$$ ----------(i)

7th term = $$A_7=a+6d=-4$$ --------(ii)

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

=> $$6d-3d=-4-11$$

=> $$3d=-15$$

=> $$d=\frac{-15}{3}=-5$$

Substituting it in equation (i), => $$a=11-3(-5)=11+15=26$$

$$\therefore$$ 15th term = $$A_{15}=a+14d$$

= $$26+14(-5)=26-70=-44$$

=> Ans - (B)

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