CBSE Class 10 Mathematics Arithmetic Progression Assignment Set 14

Question. Find the Sum of 20 terms whose first term is 8 and common difference is 5.
Answer: We are given:

• First term (\( a \)) = 8
• Common difference (\( d \)) = 5
• Number of terms (\( n \)) = 20

The sum of the first \( n \) terms of an arithmetic progression is: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Substituting the given values: \[ S_{20} = \frac{20}{2} [2(8) + (20-1)5] \] \[ S_{20} = 10 [16 + 19 \times 5] \] \[ S_{20} = 10 [16 + 95] \] \[ S_{20} = 10 \times 111 = 1110 \] Thus, the sum is \( 1110 \).

Question. Find the Sum of 50 terms whose first term is 10 and the common difference is (-2).
Answer: We are given:

• First term (\( a \)) = 10
• Common difference (\( d \)) = -2
• Number of terms (\( n \)) = 50

Using the sum formula: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Substituting the values: \[ S_{50} = \frac{50}{2} [2(10) + (50-1)(-2)] \] \[ S_{50} = 25 [20 + 49(-2)] \] \[ S_{50} = 25 [20 - 98] \] \[ S_{50} = 25 \times (-78) = -1950 \] Thus, the sum is \( -1950 \).

Question. Find the Sum of 17 terms in the following A.P.
\( 5, 1, -3, \dots \dots \dots 17\text{ terms} \)
Answer: We are given the arithmetic progression: \( 5, 1, -3, \dots \)

• First term (\( a \)) = 5
• Common difference (\( d \)) = \( 1 - 5 = -4 \)
• Number of terms (\( n \)) = 17

Using the sum formula: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Substituting the values: \[ S_{17} = \frac{17}{2} [2(5) + (17-1)(-4)] \] \[ S_{17} = \frac{17}{2} [10 + 16(-4)] \] \[ S_{17} = \frac{17}{2} [10 - 64] \] \[ S_{17} = \frac{17}{2} [-54] \] \[ S_{17} = 17 \times (-27) = -459 \] *(Note: The provided assignment answer key lists \( -4089 \), which corresponds to the sum of 47 terms: \( S_{47} = \frac{47}{2}[10 + 46(-4)] = -4089 \). This indicates a typographical error in either the question text or the key's calculation where "17" was confused with "47".)*

Question. Find the Sum of 100 terms of the following series
\( 0.9, 0.91, 0.92, \dots \dots \dots \)
Answer: We are given the arithmetic sequence: \( 0.9, 0.91, 0.92, \dots \)

• First term (\( a \)) = 0.9
• Common difference (\( d \)) = \( 0.91 - 0.9 = 0.01 \)
• Number of terms (\( n \)) = 100

Using the sum formula: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Substituting the values: \[ S_{100} = \frac{100}{2} [2(0.9) + (100-1)(0.01)] \] \[ S_{100} = 50 [1.8 + 99(0.01)] \] \[ S_{100} = 50 [1.8 + 0.99] \] \[ S_{100} = 50 \times 2.79 = 139.5 \] Thus, the sum is \( 139.5 \).

Question. Find the Sum of the following A.P.
(i) \( 2+5+8+ \dots \dots \dots +47 \)
(ii) \( 2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{6} + \dots \dots \dots + 27\frac{1}{2} \)
Answer:
(i) For the series \( 2 + 5 + 8 + \dots + 47 \):

• First term (\( a \)) = 2
• Common difference (\( d \)) = \( 5 - 2 = 3 \)
• Last term (\( a_n \)) = 47

First, find the number of terms (\( n \)): \[ a_n = a + (n-1)d \] \[ 47 = 2 + (n-1)3 \] \[ 45 = 3(n-1) \implies n - 1 = 15 \implies n = 16 \] Calculating the sum: \[ S_{16} = \frac{n}{2} [a + a_n] \] \[ S_{16} = \frac{16}{2} [2 + 47] = 8 \times 49 = 392 \] (ii) For the series \( 2\frac{1}{2} + 3\frac{1}{3} + 4\frac{1}{6} + \dots + 27\frac{1}{2} \):
Converting to improper fractions: \[ \frac{5}{2} + \frac{10}{3} + \frac{25}{6} + \dots + \frac{55}{2} \]

• First term (\( a \)) = \( \frac{5}{2} \)
• Common difference (\( d \)) = \( \frac{10}{3} - \frac{5}{2} = \frac{20 - 15}{6} = \frac{5}{6} \)
• Last term (\( a_n \)) = \( \frac{55}{2} \)

First, find the number of terms (\( n \)): \[ a_n = a + (n-1)d \] \[ \frac{55}{2} = \frac{5}{2} + (n-1)\frac{5}{6} \] \[ \frac{50}{2} = (n-1)\frac{5}{6} \] \[ 25 = (n-1)\frac{5}{6} \implies n - 1 = 30 \implies n = 31 \] Calculating the sum: \[ S_{31} = \frac{n}{2} [a + a_n] \] \[ S_{31} = \frac{31}{2} \left[\frac{5}{2} + \frac{55}{2}\right] \] \[ S_{31} = \frac{31}{2} [30] = 31 \times 15 = 465 \] Thus, the answers are: (i) \( 392 \) and (ii) \( 465 \).

Question. Find the Sum of first 18 terms of an A.P. whose first term is 5 and the common difference is 2.
Answer: We are given:

• First term (\( a \)) = 5
• Common difference (\( d \)) = 2
• Number of terms (\( n \)) = 18

Using the sum formula: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] Substituting the values: \[ S_{18} = \frac{18}{2} [2(5) + (18-1)2] \] \[ S_{18} = 9 [10 + 17 \times 2] \] \[ S_{18} = 9 [10 + 34] \] \[ S_{18} = 9 \times 44 = 396 \] Thus, the sum is \( 396 \).

Question. Find the Sum of 20 terms of the sequence whose \( n^{\text{th}} \) term is \( (3-4n) \).
Answer: The \( n^{\text{th}} \) term is: \[ a_n = 3 - 4n \] Finding the first term (\( a_1 \)) where \( n = 1 \): \[ a_1 = 3 - 4(1) = -1 \] Finding the \( 20^{\text{th}} \) term (\( a_{20} \)) where \( n = 20 \): \[ a_{20} = 3 - 4(20) = 3 - 80 = -77 \] Using the sum formula: \[ S_n = \frac{n}{2} [a_1 + a_n] \] For \( n = 20 \): \[ S_{20} = \frac{20}{2} [a_1 + a_{20}] \] \[ S_{20} = 10 [-1 + (-77)] \] \[ S_{20} = 10 \times (-78) = -780 \] Thus, the sum is \( -780 \).

Question. The first term of an A.P. is (-2) and the common difference is \( \left(-\frac{7}{2}\right) \). Find the Sum to n terms of the A.P. Hence find the Sum of 20 terms of the A.P.
Answer: We are given:

• First term (\( a \)) = -2
• Common difference (\( d \)) = \( -\frac{7}{2} \)

The sum of the first \( n \) terms is: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_n = \frac{n}{2} \left[2(-2) + (n-1)\left(-\frac{7}{2}\right)\right] \] \[ S_n = \frac{n}{2} \left[-4 - \frac{7n}{2} + \frac{7}{2}\right] \] \[ S_n = \frac{n}{2} \left[-\frac{7n}{2} - \frac{1}{2}\right] \] \[ S_n = -\frac{n(7n+1)}{4} \] To find the sum of 20 terms (\( n = 20 \)): \[ S_{20} = -\frac{20(7(20)+1)}{4} \] \[ S_{20} = -5 \times (140 + 1) \] \[ S_{20} = -5 \times 141 = -705 \] Thus, the sum to \( n \) terms is \( -\frac{n(7n+1)}{4} \) and the sum of 20 terms is \( -705 \).

Question. Find the Sum of all natural numbers between 2 and 101, which are divisible by 5.
Answer: The natural numbers strictly between 2 and 101 that are divisible by 5 form the following A.P.: \[ 5, 10, 15, \dots, 100 \]

• First term (\( a \)) = 5
• Common difference (\( d \)) = 5
• Last term (\( a_n \)) = 100

To find the number of terms (\( n \)): \[ a_n = a + (n-1)d \] \[ 100 = 5 + (n-1)5 \] \[ 95 = 5(n-1) \implies n-1 = 19 \implies n = 20 \] Calculating the sum: \[ S_{20} = \frac{n}{2} [a + a_n] \] \[ S_{20} = \frac{20}{2} [5 + 100] \] \[ S_{20} = 10 \times 105 = 1050 \] Thus, the sum is \( 1050 \).

Question. The last term of an A.P. is 252. Its 1st term and Common difference are 12 and 6. Find the Sum of A.P. by
(i) using last term
(ii) Without using last term
Verify that the Answer is Same in both cases
Answer: We are given:

• Last term (\( l \) or \( a_n \)) = 252
• First term (\( a \)) = 12
• Common difference (\( d \)) = 6

First, find the total number of terms (\( n \)): \[ a_n = a + (n-1)d \] \[ 252 = 12 + (n-1)6 \] \[ 240 = 6(n-1) \implies n-1 = 40 \implies n = 41 \] (i) Using the last term: \[ S_n = \frac{n}{2} [a + l] \] \[ S_{41} = \frac{41}{2} [12 + 252] \] \[ S_{41} = \frac{41}{2} [264] = 41 \times 132 = 5412 \] (ii) Without using the last term: \[ S_n = \frac{n}{2} [2a + (n-1)d] \] \[ S_{41} = \frac{41}{2} [2(12) + (41-1)6] \] \[ S_{41} = \frac{41}{2} [24 + 40 \times 6] \] \[ S_{41} = \frac{41}{2} [24 + 240] \] \[ S_{41} = \frac{41}{2} [264] = 41 \times 132 = 5412 \] Both methods yield the same sum of \( 5412 \), verifying the result.