Maharashtra Board Class 10 Maths Chapter 3 Arithmetic Progression Set 3.3 Solutions

Question 1. First term and common difference of an A.P. are 6 and 3 respectively; find S27.
Answer: Solution:
a = 6, d = 3, S27 = ? \[S_n = \frac{n}{2}[2a+(n-1)d]\] \[ \therefore S_{27} = \frac{27}{2}[2(6)+(27-1)3]\] \[ = \frac{27}{2}[12+(26)3]\] \[ = \frac{27}{2}[12+78]\] \[ = \frac{27}{2} \times 90\] \[ = 27 \times 45\] \[ = 1215\] In simple words: To find the sum of the first 27 terms of an A.P., we use the formula \(S_n = \frac{n}{2}[2a+(n-1)d]\) with the given first term (a) and common difference (d).

🎯 Exam Tip: Remember to correctly substitute the values of 'a', 'd', and 'n' into the sum formula and perform the calculations carefully to avoid errors.

Question 2. Find the sum of first 123 even natural numbers.
Answer: Solution:
The even natural numbers are 2, 4, 6, 8,...
The above sequence is an A.P.
∴ a = 2, d = 4 - 2 = 2, n = 123 \[Now, S_n = \frac{n}{2}[2a+(n-1)d]\] \[S_n = \frac{123}{2}[2(2)+(123-1)(2)]\] \[ = \frac{123}{2}[2(2)+122(2)]\] \[ = \frac{123}{2} \times 2[2+122]\] \[ = 123 \times 124\] \[ = 15252\] ∴ The sum of first 123 even natural numbers is 15252.
In simple words: The problem asks for the sum of the first 123 even natural numbers, which form an Arithmetic Progression with first term 2 and common difference 2. We use the sum formula for an A.P.

🎯 Exam Tip: Identify 'a', 'd', and 'n' correctly from the problem description. For even natural numbers, 'a' is 2 and 'd' is 2.

Question 3. Find the sum of all even numbers between 1 and 350.
Answer: Solution:
The even numbers between 1 and 350 are 2, 4, 6,..., 348.
The above sequence is an A.P.
∴ a = 2, d = 4 - 2 = 2, tn = 348
Since, tn = a + (n - 1)d
∴ 348 = 2 + (n - 1)2
∴ 348 - 2 = (n - 1)2
∴ 346 = (n - 1)2
\[\therefore n - 1 = \frac{346}{2}\]
∴ n - 1 = 173
∴ n = 173 + 1 = 174
\[Now, S_n = \frac{n}{2}[2a+(n-1)d]\] \[\therefore S_{174} = \frac{174}{2}[2(2)+(174-1)2]\] \[ = 87(4+173 \times 2)\] \[ = 87(4+346)\] \[ = 87 \times 350\] \[\therefore S_{174} = 30450\] ∴ The sum of all even numbers between 1 and 350 is 30450.
In simple words: First, find the number of even terms between 1 and 350 using the \(t_n\) formula, then use the sum formula \(S_n\) to calculate their total sum.

🎯 Exam Tip: For problems asking for numbers 'between' two values, make sure to exclude the endpoints if they are not explicitly part of the sequence (e.g., 'even numbers between 1 and 350' start at 2 and end at 348).

Question 4. In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Answer: Solution:
For an A.P., let a be the first term and d be the common difference.
t19 = 52, t38 = 128 ... [Given]
Since, tn = a + (n - 1)d
∴ t19 = a + (19 - 1)d
∴ 52 = a + 18d
i.e. a + 18d = 52 ...(i)
Also, t38 = a + (38 - 1)d
∴ 128 = a + 37d
i.e. a + 37d = 128 ...(ii)
Adding equations (i) and (ii), we get \[\begin{array}{r}a+18d=52\\+\quad a+37d=128\\\hline 2a+55d=180\end{array}\] ...(iii)
\[Now, S_n = \frac{n}{2}[2a+(n-1)d]\] \[\therefore S_{56} = \frac{56}{2}[2a+(56-1)d]\] \[ = 28(2a+55d)\] \[ = 28 \times 180 \quad \text{...[From (iii)]}\] \[S_{56} = 5040\] ∴ The sum of first 56 terms is 5040.
In simple words: Use the given 19th and 38th terms to form two equations and solve them to find a relationship between 'a' and 'd'. Then substitute this relationship into the sum formula \(S_{56}\).

🎯 Exam Tip: When given two terms of an A.P., set up a system of two linear equations in 'a' and 'd' and solve them. Recognize common expressions that simplify calculations, like \(2a+55d\).

Question 5. Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.
Answer: Solution:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक गतिविधि आधारित प्रश्न है जो 1 से 140 के बीच 4 से विभाज्य प्राकृतिक संख्याओं का योग ज्ञात करने की प्रक्रिया को एक फ्लोचार्ट के रूप में दर्शाता है। इसमें संख्याओं को पहचानना, 'n', 'a', 'd' का मान निकालना, और अंत में योग \(S_n\) की गणना करना शामिल है। From 1 to 140, natural numbers divisible by 4
4, 8, ....., 136
How many numbers?
∴ n = 34
n = 34, a = 4, d = 4
tn = a + (n-1)d
136 = 4 + (n-1) × 4
n = 34
\[S_n = \frac{n}{2}[2a+(n-1)d]\] \[S_{34} = \frac{34}{2}[2 \times 4 + (34-1)4] = 2380\] Sum of numbers from 1 to 140, which are divisible by 4 = 2380
In simple words: This activity involves identifying numbers divisible by 4 between 1 and 140, finding the count (n), first term (a), and common difference (d), and then applying the sum formula \(S_n\).

🎯 Exam Tip: In activity-based questions, carefully fill in the blanks using the correct formulas for arithmetic progression, ensuring each step logically follows the previous one.

Question 6. Sum of first 55 terms in an A.P. is 3300, find its 28th term.
Answer: Solution:
For an A.P., let a be the first term and d be the common difference.
S55 = 3300 ... [Given]
\[Since, S_n = \frac{n}{2}[2a+(n-1)d]\] \[\therefore S_{55} = \frac{55}{2}[2a+(55-1)d]\] \[3300 = \frac{55}{2}(2a+54d)\] \[3300 = \frac{55}{2} \times 2(a+27d)\] \[3300 = 55(a+27d)\] \[\therefore a+27d = \frac{3300}{55}\] a+27d = 60 ...(i)
Now, tn = a + (n-1)d
\[\therefore t_{28} = a+(28-1)d = a+27d\] \[\therefore t_{28} = 60 \quad \text{...[From (i)]}\] 28th term of the A.P. is 60.
In simple words: Given the sum of the first 55 terms, we use the sum formula to establish a relationship involving 'a' and 'd'. This relationship directly gives us the 28th term, as \(t_{28} = a + 27d\).

🎯 Exam Tip: Recognize that \(a+(n-1)d\) is the formula for the nth term. Sometimes, simplifying the sum formula can directly yield the desired term, avoiding the need to solve for 'a' and 'd' separately.

Question 7. In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a - d, a, a + d.)
Answer: Solution:
Let the three consecutive terms in an A.P. be
a - d, a and a + d.
According to the first condition,
a - d + a + a + d = 27
∴ 3a = 27
\[\therefore a = \frac{27}{3}\]
∴ a = 9 ....(i)
According to the second condition,
(a - d) a (a + d) = 504
∴ a(a² - d²) = 504
∴ 9(a² - d²) = 504 ...[From (i)]
∴ 9(81 - d²) = 504
\[\therefore 81 - d^2 = \frac{504}{9}\]
∴ 81 - d² = 56
∴ d² = 81 - 56
∴ d² = 25
Taking square root of both sides, we get
d = ± 5
When d = 5 and a = 9,
a - d = 9 - 5 = 4
a = 9
a + d = 9 + 5 = 14
When d = -5 and a = 9,
a - d = 9 - (-5) = 9 + 5 = 14
a = 9
a + d = 9 - 5 = 4
∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.
In simple words: By assuming the terms as \(a-d, a, a+d\), the sum condition directly gives 'a'. Then, substitute 'a' into the product condition to solve for 'd', which yields two possible sets of terms.

🎯 Exam Tip: When dealing with consecutive terms in an A.P., using \(a-d, a, a+d\) (for three terms) or \(a-3d, a-d, a+d, a+3d\) (for four terms) often simplifies calculations for sum and product conditions.

Question 8. Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14. (Assume the four consecutive terms in A.P. are a - d, a, a + d, a + 2d.)
Answer: Solution:
Let the four consecutive terms in an A.P. be
a - d, a, a + d and a + 2d.
According to the first condition,
a - d + a + a + d + a + 2d = 12
∴ 4a + 2d = 12
∴ 2(2a + d) = 12
\[\therefore 2a + d = \frac{12}{2}\]
∴ 2a + d = 6 ...(i)
According to the second condition,
a + d + a + 2d = 14
∴ 2a + 3d = 14 ...(ii)
Subtracting equation (i) from (ii), we get \[\begin{array}{r}2a+3d=14\\-\quad (2a+d=6)\\\hline 2d=8\end{array}\] \[\therefore d = \frac{8}{2}=4\] Substituting d = 4 in equation (i), we get
2a + 4 = 6
2a = 6 - 4 = 2
\[\therefore a = \frac{2}{2}=1\] a - d = 1 - 4 = -3
a = 1
a + d = 1 + 4 = 5
a + 2d = 1 + 2(4)
= 1 + 8 = 9
∴ The four consecutive terms are -3, 1, 5 and 9.
In simple words: Set up two linear equations using the given conditions (sum of all terms and sum of 3rd and 4th terms) with 'a' and 'd'. Solve these equations simultaneously to find 'a' and 'd', then determine the terms.

🎯 Exam Tip: Be careful with the chosen representation of consecutive terms. For four terms, \(a-d, a, a+d, a+2d\) works, but \(a-3d, a-d, a+d, a+3d\) might simplify the initial sum calculation further if only sum of all terms is given.

Question 9. If the 9th term of an A.P. is zero, then show that the 29th term is twice the 19th term.
Answer: Proof:
For an A.P., let a be the first term and d be the common difference.
t9 = 0 ... [Given]
Since, tn = a + (n - 1)d
∴ t9 = a + (9 - 1)d
∴ 0 = a + 8d
∴ a = -8d ...(i)
Also, t19 = a + (19 - 1)d
= a + 18d
= -8d + 18d ... [From (i)]
∴ t19 = 10d ...(ii)
and t29 = a + (29 - 1)d
= a + 28d
= -8d + 28d ... [From (i)]
∴ t29 = 20d = 2(10d)
∴ t29 = 2(t19) ... [From (ii)]
∴ The 29th term is twice the 19th term.
In simple words: Express 'a' in terms of 'd' using the given condition \(t_9 = 0\). Then, express \(t_{19}\) and \(t_{29}\) in terms of 'd' using the derived 'a', and show that \(t_{29}\) is double \(t_{19}\).

🎯 Exam Tip: Proof-based questions require clear, logical steps. Start with the given information, use formulas correctly, and substitute intermediate results to reach the desired conclusion.

Question 10. Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)
Answer: Solution:
Odd numbers from 1 to 150 are 1, 3, 5, 7,..., 149
Here, difference between any two consecutive terms is 2.
∴ It is an A.P.
∴ a = 1, d = 2
Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if
tn = 149
tn = a + (n - 1)d
∴ 149 = 1 + (n - 1)2
∴ 149 - 1 = (n - 1)2
\[\frac{148}{2} = n-1\] ∴ 74 = n - 1
∴ n = 74 + 1 = 75
ii. Now, let's find the sum of 75 numbers
i. e. 1 + 3 + 5 + 7 + ... + 149

Method I:

Here, a = 1, d = 2, n = 75
\[S_n = \frac{n}{2}[2a+(n-1)d]\] \[\therefore S_{75} = \frac{75}{2}[2(1)+(75-1)2]\] \[ = \frac{75}{2}(2+74 \times 2)\] \[ = \frac{75}{2}(2+148)\] \[ = \frac{75}{2}(150)\] \[ = 75 \times 75\] \[\therefore S_{75} = 5625\]

Method II:

Here, t1 = a = 1, tn = 149, n = 75
\[S_n = \frac{n}{2}(t_1+t_n)\] \[\therefore S_{75} = \frac{75}{2}(1+149)\] \[ = \frac{75}{2} \times 150\] \[ = 75 \times 75\] \[S_{75} = 5625\] In simple words: First, determine the total count (n) of odd numbers between 1 and 150 using the \(t_n\) formula. Then, calculate the sum of these 'n' terms using either the \(S_n\) formula with 'a', 'd', 'n' or the simplified formula using 'a', 'n', and 't_n'.

🎯 Exam Tip: When the first term, last term, and number of terms are known, the formula \(S_n = \frac{n}{2}(a+t_n)\) can be a quicker way to find the sum compared to the standard \(S_n\) formula.