Question 1 and 2 Exercise 4.2

Solutions of Question 1 and 2 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the $15$th term of the arithmetic sequence $2,5,8, \ldots$

Here $a_1=2$, $d=5-2=3$ and $n=15$. We know that: $$a_n=a_1+(n-1) d$$ This gives \begin{align}a_{15}&=2+(15-1) 3 \\ &=2+42=44 \end{align} Hence the 15th term of the given sequence is $44$.

The first term of an arithmetic sequence is 8 and the 21st is 108. Find the 7th term.

Since $a_1=8$ and $a_{21}=108$.

We know that $$a_n=a_1+(n-1) d.$$ This gives \begin{align} &a_{21}=8+(21-1) d \\ \implies &108=8+20 d\\ \implies &20 d=108-8=100 \\ \implies &d=\dfrac{100}{20}\\ \implies &d=5. \end{align} Now \begin{align}&a_7=a_1+(n-1) d \\ \implies &a_7=8+(7-1)5 \\ \implies &a_7=8+30=38 .\end{align} Hence the 7th term of the given sequence is 38.

Question 3 & 4 >