The first and fifth term of an A.P. of 40 terms are -29 and -15 respectively. Find the sum of all positive terms of this A.P

Option b: 1705. Solution: Let the first term be a = -29 and the common difference be d . We are given: t5 = a + 4d = -15 -29 + 4d = -15 4d = 14 d = 3.5 Let us find which terms are positive: tk = a + (k-1)d > 0 -29 + (k-1)3.5 > 0 (k-1)3.5 > 29 k-1 > 8.28 k > 9.28 So the first positive term is the 10th term ( t10 ): t10 = -29 + 9(3.5) = 2.5 The number of positive terms is 40 - 9 = 31 terms. The last term is t40 : t40 = a + 39d = -29 + 39(3.5) = -29 + 136.5 = 107.5 Sum of these 31 positive terms is: Spos = 31/2 [t10 + t40] Spos = 31/2 [2.5 + 107.5] = 31/2 [110] = 31 × 55 = 1705 Hence, Option B is the correct answer.

The first and fifth term of an A.P. of $40$ terms are $-29$ and $-15$ respectively. Find the sum of all positive terms of this A.P

The correct answer is Option b: 1705. Let the first term be $a = -29$ and the common difference be $d$. We are given: $$t_5 = a + 4d = -15$$ $$-29 + 4d = -15 \implies 4d = 14 \implies d = 3.5$$ Let us find which terms are positive: $$t_k = a + (k-1)d > 0$$ $$-29 + (k-1)3.5 > 0$$ $$(k-1)3.5 > 29 \implies k-1 > 8.28 \implies k > 9.28$$ So the first positive term is the $10^{\text{th}}$ term ($t_{10}$): $$t_{10} = -29 + 9(3.5) = 2.5$$ The number of positive terms is $40 - 9 = 31$ terms. The last term is $t_{40}$: $$t_{40} = a + 39d = -29 + 39(3.5) = -29 + 136.5 = 107.5$$ Sum of these $31$ positive terms is: $$S_{\text{pos}} = \frac{31}{2} [t_{10} + t_{40}]$$ $$S_{\text{pos}} = \frac{31}{2} [2.5 + 107.5] = \frac{31}{2} [110] = 31 \cdot 55 = 1705$$ Hence, **Option B** is the correct answer.

Sequence and SeriesMCQMTP June 22Question 1801 of 212

All Questions

The first and fifth term of an A.P. of $\displaystyle 40$ terms are $\displaystyle -29$ and $\displaystyle -15$ respectively. Find the sum of all positive terms of this A.P

Options

A1605

B1705

C1805

DNone of these

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 1705

All Options:

A1605
B1705
C1805
DNone of these

Detailed Solution & Explanation

Let the first term be

\displaystyle a = -29

and the common difference be

\displaystyle d

.
We are given:

t_5 = a + 4d = -15

-29 + 4d = -15 \implies 4d = 14 \implies d = 3.5

Let us find which terms are positive:

t_k = a + (k-1)d > 0

-29 + (k-1)3.5 > 0

(k-1)3.5 > 29 \implies k-1 > 8.28 \implies k > 9.28

So the first positive term is the

\displaystyle 10^{\text{th}}

term (

\displaystyle t_{10}

t_{10} = -29 + 9(3.5) = 2.5

The number of positive terms is

\displaystyle 40 - 9 = 31

terms.
The last term is

\displaystyle t_{40}

t_{40} = a + 39d = -29 + 39(3.5) = -29 + 136.5 = 107.5

Sum of these

\displaystyle 31

positive terms is:

S_{\text{pos}} = \frac{31}{2} [t_{10} + t_{40}]

S_{\text{pos}} = \frac{31}{2} [2.5 + 107.5] = \frac{31}{2} [110] = 31 \cdot 55 = 1705

Hence, **Option B** is the correct answer.

About This Chapter: Sequence and Series

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Arithmetic & Geometric Progressions

This chapter covers Arithmetic Progressions (AP) and Geometric Progressions (GP). Students learn how to find the nth term, sum of n terms, arithmetic/geometric means, and sum to infinity of a GP.

View Official ICAI Syllabus

Exam Strategy Tip

For complex 'sum of series' questions, a great hack is to substitute n = 1 and n = 2 into the question and the options to see which one matches, completely bypassing the formula.

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The first and fifth term of an A.P. of 40\displaystyle 4040 terms are −29\displaystyle -29−29 and −15\displaystyle -15−15 respectively. Find the sum of all positive terms of this A.P