4.2 Q-15

Exercise 4.2 — Question 15

Problem Statement

Find the number of terms in the arithmetic sequence if the first term is $a_1=7$, the last term is $a_n=49$, and the common difference is $d=3$.

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Key Formula

The general (nth) term of an arithmetic sequence is:

$a_n=a_1+(n-1)d$

where:

• $a_1$ = first term
• $d$ = common difference
• $n$ = number of terms
• $a_n$ = nth (last) term

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Solution

Given:

• $a_1=7$
• $a_n=49$
• $d=3$

Step 1: Substitute the known values into the formula:

$49=7+(n-1)(3)$

Step 2: Simplify:

$49-7=(n-1)(3)$

$42=3(n-1)$

Step 3: Solve for $n$:

$n-1=\frac{42}{3}=14$

$n=15$

Answer: The arithmetic sequence has $\boxed{15}$ terms.

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Verification

Check: $a_{15}=7+(15-1)(3)=7+42=49$ ✓

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Concept Summary

To identify a sequence as arithmetic, verify that the difference between consecutive terms is constant: $d=a_2-a_1=a_3-a_2=\cdots$