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Arithmetic Progressions
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Find the sum of first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n

Step-by-Step Solution

Key Concept: The given sequence is an arithmetic progression (AP) because the nth term is of the form a_n = a + (n-1)d. Identify the first term (a) and common difference (d), then apply the AP sum formula \(S_n = \frac{n}{2}[2a+(n-1)d]\) or \(S_n = \frac{n}{2}(a+l)\) where \(l\) is the last term.
1. Identify the AP parameters\
The nth term is \(a_n = 3 + 2n\).\
For \(n=1\): \(a_1 = 3 + 2\times1 = 5\). Hence the first term \(a = 5\).\
The common difference \(d\) is obtained from \(a_{n+1} - a_n\):\
\[d = (3+2(n+1)) - (3+2n) = 2\].\
So, \(d = 2\).\

2. Find the 24th (last) term\
\[l = a + (n-1)d = 5 + (24-1)\times2 = 5 + 46 = 51\].\

3. Apply the sum formula for an AP\
Using \(S_n = \frac{n}{2}(a + l)\):\
\[S_{24} = \frac{24}{2}(5 + 51) = 12 \times 56 = 672\].\
Alternatively, using \(S_n = \frac{n}{2}[2a + (n-1)d]\):\
\[S_{24} = \frac{24}{2}[2\times5 + (24-1)\times2] = 12[10 + 46] = 12 \times 56 = 672\].\

4. Result\
The sum of the first 24 terms is \(\boxed{672}\).

Correct Answer: 672
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