Question 5 and 6, Exercise 4.2

Solutions of Question 5 and 6 of Exercise 4.2 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find an arithmetic sequence for $a_{17}=-40$ and $a_{28}=-73$, find $a_{1}$ and $d$. Write first five terms of the sequence.

Solution.

The nth term of the arithmetic sequence is given as $$a_n=a_1+(n-1)d$$ Given \begin{align*} & a_{17} = -40 \\ \implies &a_1 + 16d = -40 \quad \cdots (1) \end{align*} Also \begin{align*} &a_{28}=-73\\ \implies &a_1 + 27d = -73 \quad \cdots (2) \end{align*} Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1& + 27d &= -73\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 16d &= \mathop-\limits_{+}40 \\ \hline & 11d &= -33\\ \end{array}&\\ \implies \boxed{d = -3} \quad \\ \end{align*} Putting the value $d$ in (1) \begin{align*} & a_1 +16(-3) = -40 \\ \implies & a_1 = -40+48 \\ \implies & \boxed{a_1=8} \end{align*} Now, \begin{align*} a_2 &= a_1 + d = 8 - 3 = 5 \\ a_3 &= a_2 + d = 5 - 3 = 2 \\ a_4 &= a_3 + d = 2 - 3 = -1 \\ a_5 &= a_4 + d = -1 - 3 = -4 \end{align*} Hence the first five terms are $8, 5, 2, -1, -4$ and $a_1 = 8$, $d=-3$. GOOD

The fifth term of an arithmetic sequence is 19 and 11 th term is 43. Find the first term and 87 th term.

Solution.

Given: $a_5 = 19$ and $a_{11} = 43$.

The nth term of the arithmetic sequence is given as $$a_n = a_1 + (n-1)d$$

Given \begin{align*} & a_5 = 19 \\ \implies & a_1 + 4d = 19 \quad \cdots (1) \end{align*} Also \begin{align*} & a_{11} = 43 \\ \implies & a_1 + 10d = 43 \quad \cdots (2) \end{align*}

Now, subtract equation (1) from equation (2): \begin{align*} \begin{array}{ccc} a_1 & + 10d &= 43\\ \mathop{}\limits_{-}a_1 &\mathop+\limits_{-} 4d &= \mathop-\limits_{+}19 \\ \hline & 6d &= 24\\ \end{array}&\\ \implies \boxed{d = 4} \quad \\ \end{align*}

Putting the value $d$ in (1) \begin{align*} & a_1 + 4(4) = 19 \\ \implies & a_1 = 19 - 16 \\ \implies & \boxed{a_1 = 3} \end{align*}

Now, \begin{align*} a_{87} &= a_1 + 86d \\ &= 3 + 86(4) = 347 \end{align*} Hence the $a_1 = 3$ and $a_{87} = 347$. GOOD

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