Question 3 and 4 Exercise 4.2

Solutions of Question 3 and 4 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 numbers of terms in arithmetic progression $6,9,12, \ldots, 78$.

Here $a_1=6$ and $d=9-6=3$ and $a_n=78$.
We know that $$a_n=a_1+(n-1) d$$ This gives \begin{align}&78=6+(n-1) 3 \\ \implies &3(n-1)=78-6 \\ \implies &n-1=\dfrac{72}{3} \\ \implies &n=24+1=25.\end{align} Thus, the number of terms in given progression are $25$.

The $n$th term of sequence is given by $a_n=2n+7$. Show that it is an arithmetic progression. Also find its 7th term.

Given that $$a_n=2 n+7. --- (1)$$ Then \begin{align}a_{n+1}=2(n+1)+7=2 n+9.\end{align} Take \begin{align} d&=a_{n+1}-a_n\\ &=2n+9-2n-7=2\end{align} This gives every two terms of the sequence has same distance $2$, hence it is an A.P.

Putting $n=7$ in (1), we get $$a_7=2(7)+7=14+7=21.$$ Hence the 7th term of given AP is 21.

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