1.5 + 2.8 + 3.11 + ... n terms =

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q3 – AP-EAMCET 2026

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Home Maths Sequences and SeriesQuestion #388

1.5 + 2.8 + 3.11 + \dots n terms =

Clue 1 of 3: Series Interpretation

When terms are presented as 'A.B', it often implies a product of two numbers, not a decimal. Look for patterns in the first and second parts of each term.

🧠 Full Solution Path

Step 1: Interpret the Series Notation and Identify General Term✦ Active

The series is given as $1.5 + 2.8 + 3.11 + \dots n terms$ . In such problems, the notation 'A.B' typically implies a product of two numbers rather than a decimal. Let's interpret the terms as products:

T_{1} = 1 \times 5 = 5

T_{2} = 2 \times 8 = 16

T_{3} = 3 \times 11 = 33

Now, let's identify the pattern for the two sequences being multiplied:

The first sequence is $1, 2, 3, \dots, k$ . The $k$ -th term of this sequence is $k$ .

The second sequence is $5, 8, 11, \dots$ . This is an arithmetic progression (AP) with first term $a = 5$ and common difference $d = 8 - 5 = 3$ . The $k$ -th term of an AP is given by $a_{k} = a + (k - 1) d$ . So, for this sequence, the $k$ -th term is $5 + (k - 1) 3 = 5 + 3 k - 3 = 3 k + 2$ .

Therefore, the general $k$ -th term ( $T_{k}$ ) of the given series is the product of the $k$ -th terms of these two sequences:

T_{k} = k (3 k + 2) = 3 k^{2} + 2 k

💡 Teacher's Secret Hint

Always confirm the interpretation of the series notation. If it were actual decimals, the differences between terms ( $1.3, 0.31$ ) would not lead to the polynomial forms seen in the options, which usually arise from sums of $k^{2}$ or $k$ terms.

Step 2: Apply Summation Formulas○ Expand

To find the sum of $n$ terms, $S_{n}$ , we need to sum the general term $T_{k}$ from $k = 1$ to $n$ :

S_{n} = \sum_{k = 1}^{n} T_{k} = \sum_{k = 1}^{n} (3 k^{2} + 2 k)

S_{n} = 3 \sum_{k = 1}^{n} k^{2} + 2 \sum_{k = 1}^{n} k

We use the standard summation formulas:

\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}

\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}

💡 Teacher's Secret Hint

Memorizing common summation formulas for $\sum k$ , $\sum k^{2}$ , and $\sum k^{3}$ is crucial for quickly solving these types of series problems. Be careful with coefficients.

Step 3: Substitute and Simplify○ Expand

Substitute the summation formulas into the expression for $S_{n}$ :

S_{n} = 3 (\frac{n (n + 1) (2 n + 1)}{6}) + 2 (\frac{n (n + 1)}{2})

Simplify the expression:

S_{n} = \frac{n (n + 1) (2 n + 1)}{2} + n (n + 1)

Factor out the common term $\frac{n (n + 1)}{2}$ :

S_{n} = \frac{n (n + 1)}{2} ((2 n + 1) + 2)

S_{n} = \frac{n (n + 1)}{2} (2 n + 1 + 2)

S_{n} = \frac{n (n + 1) (2 n + 3)}{2}

💡 Teacher's Secret Hint

Always look for common factors to simplify the expression at each step. This makes the algebra less cumbersome and helps in matching with the given options.

Step 4: Match with Options○ Expand

The derived sum $S_{n} = \frac{n (n + 1) (2 n + 3)}{2}$ matches option 1.

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Maths Question 3 – AP-EAMCET 2026

Clue 1 of 3: Series Interpretation

🧠 Full Solution Path