Question 22

loading...

Prev/Next links

Jambmaths question: 

The 3rd term of an arithmetic progression is  – 9 and the 7th  term is –29. Find the 10th term of the progression

Jamb Maths Solution: 

$\begin{align}  & {{T}_{3}}=-9 \\ & {{T}_{7}}=-29 \\ & {{T}_{n}}=a+(n-1)d \\ & {{T}_{3}}=a+(3-1)d=-9 \\ & a+2d=-9---(i) \\ & {{T}_{7}}=a+(7-1)d=-29 \\ & a+6d=-29---(ii) \\ & \text{subtract (2) from (1)} \\ & -4d=20 \\ & \text{   }d=-5 \\ & \text{substitute }-5\text{ for }d\text{ in equation (i)} \\ & a+2(-5)=-9 \\ & a=1 \\ & {{T}_{10}}=1+(10-1)(-5)=1-45=-44 \\\end{align}$

Jamb Maths Topic: 
Year of Exam: