CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 12

Question. In an AP, if \( a = 4 \), \( n = 7 \) and \( a_n = 4 \), then the value of ‘d’ is 
(a) 0
(b) 1
(c) 3
(d) 2
Answer: (a) 0
Answer: Given: \( a = 4 \), \( n = 7 \) and \( a_n = 4 \), then
\( a_n = a + (n - 1)d \)

\( \implies 4 = 4 + (7 - 1)d \)

\( \implies 4 - 4 = 6d \)

\( \implies 6d = 0 \)

\( \implies d = 0 \)

Question. The next two terms of the AP : \( k, 2k + 1, 3k + 2, 4k + 3, ………… \) are 
(a) \( 5k + 4 \) and \( 6k + 5 \)
(b) \( 4k + 4 \) and \( 4k + 5 \)
(c) \( 5k + 5 \) and \( 6k + 6 \)
(d) \( 5k \) and \( 6k \)
Answer: (a) \( 5k + 4 \) and \( 6k + 5 \)
Answer: Given: \( k, 2k + 1, 3k + 2, 4k + 3, \dots \)
Here \( d = 2k + 1 - k = k + 1 \)
Therefore, the next two terms are
\( 4k + 3 + k + 1 = 5k + 4 \) and \( 5k + 4 + k + 1 = 6k + 5 \)

Question. The common difference of the A.P. can be 
(a) only negative
(b) only zero
(c) positive, negative or zero
(d) only positive
Answer: (c) positive, negative or zero
Answer: The common difference of the A.P. can be positive, e.g. 1, 2, 3, 4 ..... d is +ve and series is increasing negative e.g 4, 3, 2, 1 ...... d is - ve and series is decreasing or zero also and the AP becomes constant e.g 4, 4, 4, 4 .......

Question. The 7th term from the end of the A.P. – 11, – 8, – 5, ……., 49 is 
(a) 28
(b) 31
(c) -11
(d) -8
Answer: (b) 31
Answer: Reversing the given A.P., we have
\( 49, 46, 43, \dots, -11 \)
Here, \( a = 49 \), \( d = 46 - 49 = -3 \) and \( n = 7 \)
\( \therefore a_n = a + (n - 1)d \)

\( \implies a_7 = 49 + (7 - 1) \times (-3) \)

\( \implies = 49 + 6 \times (-3) \)

\( \implies a_7 = 49 - 18 = 31 \)

Question. The common difference of the A.P whose \( a_n = -3n + 7 \) is 
(a) 3
(b) 1
(c) -3
(d) 2
Answer: (c) – 3
Answer: Given: \( a_n = -3n + 7 \)
Putting \( n = 1, 2, 3 \), we get
\( a = -3 \times 1 + 7 = -3 + 7 = 4 \)
\( a_2 = -3 \times 2 + 7 = -6 + 7 = 1 \)
\( a_3 = -3 \times 3 + 7 = -9 + 7 = -2 \)
\( \therefore \) Common difference \( (d) = a_2 - a = 1 - 4 = -3 \)

Question. If 5 times the \( 5^{th} \) term of an AP is equal to 10 times the \( 10^{th} \) term, show that its \( 15^{th} \) term is zero. 
Answer: Let \( 1^{st} \) term = \( a \) and common difference = \( d \).
\( a_5 = a + 4d \), \( a_{10} = a + 9d \)
According to the question, \( 5 \times a_5 = 10 \times a_{10} \)

\( \implies 5(a + 4d) = 10(a + 9d) \)

\( \implies 5a + 20d = 10a + 90d \)

\( \implies a = -14d \)
Now, \( a_{15} = a + 14d \)

\( \implies a_{15} = -14d + 14d = 0 \).

Question. Write the first term a and the common difference d of A.P. -1.1, - 3.1, -5.1, - 7.1,... 
Answer: First term \( (a) = -1.1 \)
We know that common difference is difference between any two consecutive terms of an A.P.
So, common difference \( (d) = (-3.1) - (-1.1) = -3.1 + 1.1 = -2 \)

Question. For what value of n are the \( n^{th} \) term of the following two AP's are same 13, 19, 25, .... and 69, 68, 67 .... 
Answer: \( n^{th} \) term of 13, 19, 25, ............ = \( n^{th} \) term of 69, 68, 67, .........
\( 13 + (n - 1) 6 = 69 + (n - 1) (-1) \)
\( 13 + 6n - 6 = 69 - n + 1 \)
\( n + 6n = 70 - 7 \)
\( 7n = 63 \)
\( n = 9 \)
Therefore, \( n = 9 \)

Question. Find k, if the given value of x is the \( k^{th} \) term of the given AP \( 5 \frac{1}{2}, 11, 16 \frac{1}{2}, 22, ..., x = 550 \). 
Answer: \( a = 5 \frac{1}{2} = \frac{11}{2} \), \( d = a_2 - a_1 = 11 - \frac{11}{2} = \frac{11}{2} \) and \( x = 550 \)
A.T.Q., \( a_k = x \)

\( \implies a + (k - 1)d = 550 \)

\( \implies \frac{11}{2} + (k - 1) \frac{11}{2} = 550 \)

\( \implies \frac{11}{2} + \frac{11}{2} k - \frac{11}{2} = 550 \)

\( \implies \frac{11}{2} k = 550 \)

\( \implies k = \frac{550 \times 2}{11} = 100 \)

Question. Find the 6th term from the end of the A.P. 17, 14, 11, ..., - 40 
Answer: A.P. is 17, 14, 11, ..., - 40
We have,
\( l = \text{Last term} = -40 \), \( a = 17 \) and, \( d = \text{Common difference} = 14 - 17 = -3 \)
\( \therefore 6^{th} \) term from the end = \( l - (n - 1)d \)
\( = l - (6 - 1)d \)
\( = -40 - 5 \times (-3) \)
\( = -40 + 15 = -25 \)
So, \( 6^{th} \) term of given A.P. is -25.

Question. How many terms of the AP 17, 15, 13, 11,... must be added to get the sum 72? 
Answer: Given A.P. is 17, 15, 13, 11........
Here, \( 1^{st} \) term \( (a) = 17 \) and common difference \( (d) = (15 - 17) = -2 \)
Let the sum of n terms be 72. Then,
\( S_n = 72 \)

\( \implies \frac{n}{2} \cdot \{2a + (n - 1)d\} = 72 \)

\( \implies n \{2 \times 17 + (n - 1)(-2)\} = 144 \)

\( \implies n(36 - 2n) = 144 \)

\( \implies 2n^2 - 36n + 144 = 0 \)

\( \implies n^2 - 18n + 72 = 0 \)

\( \implies n^2 - 12n - 6n + 72 = 0 \)

\( \implies n(n - 12) - 6(n - 12) = 0 \)

\( \implies (n - 12)(n - 6) = 0 \)

\( \implies n = 6 \) or \( n = 12 \).
\( \therefore \) sum of first 6 terms = sum of first 12 terms = 72. This means that the sum of all terms from 7th to 12th is zero.

Question. Find n. Given a = first term = -18.9, d = common difference = 2.5, \( a_n \) = the nth term = 3.6, n = ? 
Answer: \( a_n = a + (n - 1)d \)

\( \implies 3.6 = - 18.9 + (n - 1)(2.5) \)

\( \implies 3.6 + 18.9 = (n - 1)(2.5) \)

\( \implies 22.5 = (n - 1)(2.5) \)

\( \implies n - 1 = \frac{22.5}{2.5} = 9 \)

\( \implies n = 10 \)

Question. Find the number of terms in each of the following APs. 18, \( 15 \frac{1}{2} \), 13, ….., – 47. 
Answer: Here, \( a = 18 \)
\( d = 15 \frac{1}{2} - 18 = \frac{31}{2} - 18 = -\frac{5}{2} \)
\( a_n = -47 \)
Let the number of terms be n.
Then,
\( a_n = -47 \)

\( \implies a + (n - 1)d = -47 \)

\( \implies 18 + (n - 1) \left( -\frac{5}{2} \right) = -47 \)

\( \implies -\frac{5}{2} (n - 1) = -47 - 18 \)

\( \implies -\frac{5}{2} (n - 1) = -65 \)

\( \implies \frac{5}{2} (n - 1) = 65 \)

\( \implies n - 1 = \frac{65 \times 2}{5} \)

\( \implies n - 1 = 26 \)

\( \implies n = 26 + 1 \)

\( \implies n = 27 \)
Hence, the number of terms of the given AP is 27.

Question. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find the common difference and the number of terms. 
Answer: Let the given AP contains n terms.
First term, \( a = 5 \)
Last term, \( l = 45 \)
\( S_n = 400 \)

\( \implies \frac{n}{2} [a + l] = 400 \)

\( \implies \frac{n}{2} [5 + 45] = 400 \)

\( \implies n \times 25 = 400 \)

\( \implies n \times 50 = 800 \)

\( \implies n = 16 \)
Thus, the given AP contains 16 terms.
Let d be the common difference of the given AP.
then,
\( T_{16} = 45 \)

\( \implies a + 15d = 45 \)

\( \implies 5 + 15d = 45 \)

\( \implies 15d = 40 \)

\( \implies d = \frac{40}{15} = \frac{8}{3} \).
Therefore, common difference of the given AP is \( \frac{8}{3} \).

Question. The 14th term of an A.P. is twice its 8th term. If the 6th term is -8, then find the sum of its first 20 terms. 
Answer: Let first term be a and common difference be d.
Here, \( a_{14} = 2a_8 \)
\( a + 13d = 2(a + 7d) \)
\( a + 13d = 2a + 14d \)
\( a = - d \) ...(i)
\( a_6 = - 8 \)
\( a + 5d = -8 \) ... (ii)
Putting the value of a from (i) in (ii), we get
\( -d + 5d = -8 \)
\( 4d = -8 \)
\( d = -2 \)
Put \( d = -2 \) in (i)
\( a = -(-2) \)
\( a = 2 \)
So \( a = 2, d = - 2 \)
\( S_{20} = \frac{20}{2} [2 \times 2 + (20 - 1)(-2)] \)
\( = 10[4 + 19 \times (-2)] \)
\( = 10(4 - 38) \)
\( = 10 \times (-34) \)
= - 340. Which is the required sum of first 20 terms.

Question. Find the 6th term from end of the AP 17, 14, 11, …, –40. 
Answer: The given AP is 17, 14, 11, ......., – 40
Here, \( a = 17 \)
\( d = 14 - 17 = -3 \)
\( l = -40 \)
Let there be n terms between in the given AP
Then, \( n^{th} \) term = -40

\( \implies a + (n - 1)d = -40 \)

\( \implies 17 + (n - 1)(-3) = -40 \)

\( \implies (n - 1)(-3) = -40 - 17 \)

\( \implies (n - 1)(-3) = -57 \)

\( \implies n - 1 = 19 \)

\( \implies n = 19 + 1 \)

\( \implies n = 20 \)
Hence, there are 20 terms in the given AP.
Now, 6th term from the end
= (20 - 6 + 1)th term from the beginning
= 15th term from the beginning
\( = a + (15 - 1)d \)
\( = 17 + 14(-3) \)
\( = 17 - 42 \)
\( = - 25 \)
Hence, the 6th term from the end of the given AP is -25.

Question. The houses of a row in a colony are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find the value of x. 
Answer: According to the question, we have to find the value of x.
We are given an AP, namely 1, 2, 3, ..., (x -1), x, (x + 1), ..., 49
such that \( 1 + 2 + 3 + ... + (x - 1) = (x + 1) + (x + 2) + ... + 49 \).
Thus, we have \( S_{x-1} = S_{49} - S_x \) ... (i)
Using the formula, \( S_n = \frac{n}{2}(a + l) \) in (i), we have,
\( \frac{(x-1)}{2} \cdot \{1 + (x - 1)\} = \frac{49}{2} \cdot (1 + 49) - \frac{x}{2} \cdot (1 + x) \)

\( \implies \frac{x(x-1)}{2} + \frac{x(x+1)}{2} = 1225 \)

\( \implies 2x^2 = 2450 \)

\( \implies x^2 = 1225 \)

\( \implies x = \sqrt{1225} = 35 \)
Hence, x = 35.

Question. The sum of first n terms of an A.P. is \( 3n^2 + 4n \). Find the \( 25^{th} \) term of this A.P. 
Answer: According to the question,
Sum of n terms of the A.P. \( S_n = 3n^2 + 4n \)
\( S_1 = 3 \times 1^2 + 4 \times 1 = 7 = t_1 \) ....(i)
\( S_2 = 3 \times 2^2 + 4 \times 2 = 20 = t_1 + t_2 \) ....(i)
\( S_3 = 3 \times 3^2 + 4 \times 3 = 39 = t_1 + t_2 + t_3 \) .... (iii)
From (i), (ii), (iii)
\( t_1 = 7 \), \( t_2 = 13 \), \( t_3 = 19 \)
Common difference, \( d = 13 - 7 = 6 \)
\( 25^{th} \) of the term of this A.P., \( t_{25} = 7 + (25 - 1)6 \)
\( = 7 + 144 = 151 \)
\( \therefore \) The \( 25^{th} \) term of the A.P. is 151.

Question. If sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms. 
Answer: Consider the A.P. whose first term and common difference are ‘a’ and ‘d’ respectively.
If sum of first 6 terms of an A.P. is 36.
\( S_6 = 36 \)
\( \therefore \frac{6}{2} [2a + (6 – 1)d] = 36 \cdot \because S_n = \frac{n}{2} [2a + (n – 1)d] \)

\( \implies 3[2a + 5d] = 36 \)

\( \implies 2a + 5d = \frac{36}{3} \)

\( \implies 2a + 5d = 12 \) …(i)
If sum of first 16 terms is 256,
So, \( S_{16} = 256 \)

\( \implies \frac{16}{2} [2a + (16 – 1)d] = 256 \)

\( \implies 8[2a + 15d] = 256 \)

\( \implies 2a + 15d = \frac{256}{8} \)

\( \implies 2a + 15d = 32 \) …(ii)
Subtracting (i) from (ii), we get
\( 2a + 15d = 32 \)
\( 2a + 5d = 12 \)
– – –
\( 10d = 20 \)

\( \implies d = 2 \)
Now, \( 2a + 5d = 12 \) [From (i)]

\( \implies 2a + 5(2) = 12 \)

\( \implies 2a + 10 = 12 \)

\( \implies 2a = 12 – 10 \)

\( \implies a = \frac{2}{2} \)

\( \implies a = 1 \)
Hence, \( a = 1 \) and \( d = 2 \)
So, \( S_{10} = \frac{10}{2} [2a + (10 – 1)d] \)
\( = 5[2(1) + 9(2)] \)
\( = 5[2 + 18] \)
\( = 5[20] \)
\( = 100 \)
\( \therefore S_{10} = 100 \)
Hence, the sum of first 10 terms is 100.

Question. Let the first term and the common difference of the AP be a and d respectively. According to the question, Third term + seventh term = 6 and their product is 8. Find the sum of the first sixteen terms of the AP. 
Answer: Third term + seventh term = 6

\( \implies [a + (3 - 1)d] + [a + (7 - 1)d] = 6 \)

\( \implies (a + 2d) + (a + 6d) = 6 \)

\( \implies 2a + 8d = 6 \)

\( \implies a + 4d = 3 \) ..... (1)
Dividing throughout by 2 &
(third term) (seventh term) = 8

\( \implies (a + 2d) (a + 6d) = 8 \)

\( \implies (a + 4d - 2d) (a + 4d + 2d) = 8 \)

\( \implies (3 - 2d) (3 + 2d) = 8 \)

\( \implies 9 - 4d^2 = 8 \)

\( \implies 4d^2 = 1 \implies d^2 = \frac{1}{4} \implies d = \pm \frac{1}{2} \)
Case I, when \( d = \frac{1}{2} \)
Then from (1), \( a + 4 \left( \frac{1}{2} \right) = 3 \)

\( \implies a + 2 = 3 \implies a = 3 - 2 \implies a = 1 \)
\( \therefore \) Sum of first sixteen terms of the AP = \( S_{16} \)
\( = \frac{16}{2} [2a + (16 - 1)d] \cdot \because S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( = 8[2a + 15d] \)
\( = 8[2(1) + 15 \left( \frac{1}{2} \right)] \)
\( = 8[12 + \frac{15}{2}] \)
\( = 8 \left[ \frac{19}{2} \right] \)
\( = 4 \times 19 = 76 \)
Case II. When \( d = - \frac{1}{2} \)
Then from (1),
\( a + 4 \left( - \frac{1}{2} \right) = 3 \)

\( \implies a - 2 = 3 \implies a = 3 + 2 \implies a = 5 \)
\( \therefore \) Sum of first sixteen terms of the AP = \( S_{16} \)
\( = \frac{16}{2} [2a + (16 - 1)d] \cdot \because S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( = 8[2a + 15d] = 8 \left[ 2(5) + 15 \left( - \frac{1}{2} \right) \right] = 8 \left[ 10 - \frac{15}{2} \right] = 8 \left[ \frac{5}{2} \right] = 20 \)

Question. Find the sum of first‘n’ terms of an A.P.? 
(a) \( 7n – 8 \)
(b) \( S = \frac{n}{2} [2a + (n - 1)d] \)
(c) \( 2n + 3 \)
(d) \( n^2 + 2 \)
Answer: (b) \( S = \frac{n}{2} [2a + (n - 1)d] \)
Explanation: let \( a = 1^{st} \) term, \( d = \text{common difference} \), \( S_n = \text{sum of } 1^{st} \text{ n terms of an AP} \)
then \( S_n = (a) + (a + d) + (a + 2d) + \dots + \{a + (n - 3)d\} + \{a + (n - 2)d\} + \{a + (n - 1)d\} \) .......... (i)
Now Rewrite \( S_n \) as follows
\( S_n = \{a + (n - 1)d\} + \{a + (n - 2)d\} + \{a + (n - 3)d\} + \dots + (a + 3d) + (a + 2d) + (a + d) + a \) ............. (ii)
adding the terms (i) and (ii) vertically
adding \( 1^{st} \) term of both we get \( (a) + \{a + (n - 1)d\} = 2a + (n - 1)d \)
adding \( 2^{nd} \) term of both \( (a + d) + \{a + (n - 2)d\} = 2a + (n - 1)d \)
adding \( 3^{rd} \) terms of both \( (a + 2d) + \{a + (n - 3)d\} = 2a + (n - 1)d \)
since there are n terms in each of the equations (i) and (ii), adding both the equations we get
\( 2S_n = n\{2a + (n - 1)d\} \)

\( \implies \) \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \).

Question. The sum of first 24 terms of the list of numbers whose nth term is given by \( a_n = 3 + 2n \) is 
(a) 680
(b) 672
(c) 640
(d) 600
Answer: (b) 672
Explanation: Given: \( a_n = 3 + 2n \)
\( \therefore a_1 = 3 + 2 \times 1 = 3 + 2 = 5 \)
\( a_2 = 3 + 2 \times 2 = 3 + 4 = 7 \)
\( \therefore d = a_2 - a_1 = 7 - 5 = 2 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies \) \( S_{24} = \frac{24}{2} [2 \times 5 + (24 - 1)2] \)

\( \implies \) \( S_{24} = 12 [10 + 23 \times 2] = 12 [10 + 46] = 672 \)

Question. The 17th term of an AP exceeds its 10th term by 7, then the common difference is 
(a) -1
(b) 1
(c) 2
(d) 0
Answer: (b) 1
Explanation: According to question, Given that the 17th term of an A.P exceeds its 10th term by 7.
d = ?

\( \implies \) \( a + 16d = a + 9d + 7 \)

\( \implies \) \( 16d - 9d = 7 \)

\( \implies \) \( 7d = 7 \)

\( \implies \) \( d = \frac{7}{7} = 1 \)
Therefore, common difference = 1.

Question. The first term of an AP is 5, the last term is 45 and the sum is 400. The number of terms is 
(a) 16
(b) 20
(c) 17
(d) 18
Answer: (a) 16
Explanation: Given: \( a = 5, l = 45, S_n = 400 \)
\( \therefore S_n = \frac{n}{2} (a + l) \)

\( \implies \) \( 400 = \frac{n}{2} (5 + 45) \)

\( \implies \) \( 800 = n \times 50 \)

\( \implies \) \( n = 16 \)

Question. If a, b and c are in A. P., then the value of \( \frac{a-b}{b-c} \) is 
(a) \( \frac{a}{b} \)
(b) 1
(c) \( \frac{c}{a} \)
(d) \( \frac{b}{c} \)
Answer: (b) 1
Explanation: If a, b and c are in A.P.,
\( b - a = c - b \)
\( -(a - b) = -(b - c) \)
\( a - b = b - c \)
dividing both sides by \( (b - c) \)
\( \frac{a-b}{b-c} = \frac{b-c}{b-c} \)

\( \implies \) \( \frac{a-b}{b-c} = 1 \)

Question. Find the common difference of the A.P. and write the next two terms of A.P. 119, 136, 153, 170,..... 
Answer: Given A.P is 119, 136, 153, 170........
We know that common difference is difference between any consecutive terms of an A.P.
So, common difference = 136 - 119 = 17
\( 5^{th} \) term = 170 + 17 = 187 (\( a_5 = a + 4d \))
\( 6^{th} \) term = 187 + 17 = 204. (\( a_6 = a + 5d \))

Question. lf 2x, x + 10, 3x + 2 are in A.P., find the value of x. 
Answer: we have to find the value of x.
Since, 2x, x + 10, 3x + 2 are in A.P. therefore \( 2 (x + 10) = 2x + 3x + 2 \)

\( \implies \) \( 2x + 20 = 5x + 2 \)

\( \implies \) \( 3x = 18 \)

\( \implies \) \( x = 6 \)

Question. Find 7th term from the end of the AP : 7, 10, 13,...., 184. 
Answer: Given, AP is 7, 10, 13,...., 184.
we have to find 7th term from the end
reversing the AP, 184,....., 13, 10, 7.
now, d = common difference = 7 - 10 = -3
\( \therefore 7^{th} \) term from the beginning of reversed AP = \( a + (7 - 1)d = a + 6d \)
= \( 184 + (6 \times (-3)) \)
= 184 - 18 = 166

Question. Find k, if the given value of x is the kth term of the given AP 25, 50, 75, 100, ..., x = 1000. 
Answer: a = 25, d = 50 - 25 = 25, x = 1000
A.T.Q., \( a_k = x \)

\( \implies \) \( a + (k - 1)d = 1000 \)

\( \implies \) \( 25 + (k - 1)25 = 1000 \)

\( \implies \) \( (k - 1)25 = 975 \)

\( \implies \) \( k - 1 = \frac{975}{25} \)

\( \implies \) \( k - 1 = 39 \)

\( \implies \) \( k = 40 \)

Question. Find the 11th term from the end of the AP 10, 7, 4, ...., -62. 
Answer: We have a = 10, d = (7 - 10) = -3, l = -62 and n = 11.
11th term from the end = \( [l - (n - 1) \times d] \)
= \( \{-62 - (11 - 1) \times (-3)\} \)
= \( (-62 + 30) = -32 \).
Hence, the 11th term from the end of the given AP is -32.

Question. Find the common difference d and write three more terms. \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \) 
Answer: \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
\( a_2 - a_1 = \sqrt{8} - \sqrt{2} = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
\( a_3 - a_2 = \sqrt{18} - \sqrt{8} = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \)
\( a_4 - a_3 = \sqrt{32} - \sqrt{18} = 4\sqrt{2} - 3\sqrt{2} = \sqrt{2} \)
i.e. \( a_{k+1} - a_k \) is the same every time.
So, the given list of numbers forms an AP with the common difference \( d = \sqrt{2} \).
The next three terms are:
\( \sqrt{32} + \sqrt{2} = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50} \)
\( 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72} \)
and \( 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98} \)

Question. The house of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the following it. Find this value of x. 
Answer: The consecutive numbers on the houses of a row are 1, 2, 3, ..., 49
Clearly this list of number forming an AP. Here, a = 1, d = 2 - 1 = 1
\( S_{x-1} = S_{49} - S_x \)

\( \implies \) \( \frac{x-1}{2} [2a + (x - 1 - 1)d] = \frac{49}{2} [2a + (49 - 1)d] - \frac{x}{2} [2a + (x - 1)d] \)

\( \implies \) \( S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies \) \( \frac{x-1}{2} [2(1) + (x - 2)(1)] = \frac{49}{2} [2(1) + (48)(1)] - \frac{x}{2} [2(1) + (x - 1)(1)] \)

\( \implies \) \( \frac{x-1}{2} [x] = 1225 - \frac{x(x+1)}{2} \)

\( \implies \) \( \frac{(x-1)(x)}{2} + \frac{x(x+1)}{2} = 1225 \)

\( \implies \) \( \frac{x}{2} (x - 1 + x + 1) = 1225 \)

\( \implies \) \( x^2 = 1225 \)

\( \implies \) \( x = \sqrt{1225} \)

\( \implies \) \( x = 35 \)
Hence, the required value of x is 35.

Question. Divide 24 in three parts such that they are in AP and their product is 440. 
Answer: Let the required numbers in A.P.are (a - d), a and (a + d).
Sum of these numbers = \( (a - d) + a + (a + d) = 3a \)
Product of these numbers = \( (a - d) \times a \times (a + d) = a(a^2 - d^2) \)
But given, sum = 24 and product = 440

\( \implies \) \( 3a = 24 \implies a = 8 \)
and \( a(a^2 - d^2) = 8(64 - d^2) = 440 \) [\( \because a = 8 \)]
Or, \( 64 - d^2 = 55 \)
Or, \( d^2 = 64 - 55 \)

\( \implies \) \( d^2 = 9 \)

\( \implies \) \( d = \pm 3 \)
When a = 8 and d = 3, The required numbers are (5, 8, 11).
When a = 8 and d = -3, The required numbers are (11, 8, 5).

Question. Determine an A.P. whose third term is 9 and when fifth term is subtracted from 8th term, we get 6. 
Answer: Let the first term be a and the common difference be d.
\( a_n = a + (n - 1)d \)
Here given, \( a_3 = 9 \)
or, \( a + 2d = 9 \) ....(i)
\( a_8 - a_5 = 6 \)
or, \( (a + 7d) - (a + 4d) = 6 \)
\( a + 7d - a - 4d = 6 \)
or, \( 3d = 6 \)
or, \( d = 2 \) ....(ii)
Substituting this value of d from (ii) in (i), we get
or, \( a + 2(2) = 9 \)
or, \( a + 4 = 9 \)
or \( a = 9 - 4 \)
or, \( a = 5 \)
a = 5 and d = 2. So, A.P. is 5,7,9,11,....

Question. Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22. 
Answer: Given \( a_6 = 12 \)

\( \implies \) \( a + (6 - 1)d = 12 \)

\( \implies \) \( a + 5d = 12 \) ............(i)
and, \( a_8 = 22 \)

\( \implies \) \( a + (8 - 1)d = 22 \)

\( \implies \) \( a + 7d = 22 \) ............(ii)
Subtracting equation (i) from (ii), we get
\( (a + 7d) - (a + 5d) = 22 - 12 \)

\( \implies \) \( a + 7d - a - 5d = 10 \)

\( \implies \) \( 2d = 10 \)

\( \implies \) \( d = \frac{10}{2} = 5 \)
Using value of d in equation (i), we get
\( a + 5 \times 5 = 12 \)

\( \implies \) \( a = 12 - 25 = -13 \)
Second term(\( a_2 \)) = \( a + (2 - 1)d = -13 + 1(5) = -13 + 5 = -8 \)
\( n^{th} \) term(\( a_n \)) = \( a + (n - 1)d \)
= \( -13 + (n - 1)(5) \)
= \( 5n - 18 \)

Question. If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term. 
Answer: We have,
\( a_9 = 0 \)

\( \implies \) \( a + (9 - 1)d = 0 \)

\( \implies \) \( a + 8d = 0 \)

\( \implies \) \( a = -8d \)
To prove: \( a_{29} = 2a_{19} \)
Proof: \( LHS = a_{29} = a + (29 - 1)d = a + 28d = -8d + 28d = 20d \)
\( RHS = 2a_{19} = 2[a + (19 - 1)d] = 2[-8d + 18d] = 2 \times 10d = 20d \)
\( \therefore LHS = RHS \)
Hence, 29th term is double the 19th term.

Question. If the sum of Rs 1890 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 50 less than its preceding prize. Then find the value of each of the prizes. 
Answer: Let \( 1^{st} \) prize be Rs x.
The series in A.P. is x, x - 50, x - 100, x - 150,........
Where a = x, d = - 50, \( S_n = 1890, n = 7 \).
As we know that \( S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies \) \( \frac{7}{2} [2x + (6)(-50)] = 1890 \)

\( \implies \) \( \frac{7}{2} [2x - 300] = 1890 \)

\( \implies \) \( 2x - 300 = 1890 (2/7) \)

\( \implies \) \( 2x = 540 + 300 \)

\( \implies \) \( x = \frac{840}{2} = 420 \)
The prizes are: Rs 420, Rs 370, Rs 320, Rs 270, Rs 220, Rs 170, Rs 120.

Question. Let a sequence be defined by \( a_1 = 1, a_2 = 1 \) and, \( a_n = a_{n-1} + a_{n-2} \) for all n > 2. Find \( \frac{a_{n+1}}{a_n} \) for n = 1,2,3,4. 
Answer: Given \( a_1 = 1 \) and \( a_2 = 1 \) and \( a_n = a_{n-1} + a_{n-2} \)
So \( a_3 = a_2 + a_1 = 1 + 1 = 2 \)
\( a_4 = a_3 + a_2 = 2 + 1 = 3 \)
\( a_5 = a_4 + a_3 = 3 + 2 = 5 \)
Now putting n = 1,2,3 and 4 in \( a_{n+1}/a_n \) we get
\( \frac{a_2}{a_1} = \frac{1}{1} = 1 \)
\( \frac{a_3}{a_2} = \frac{2}{1} = 2 \)
\( \frac{a_4}{a_3} = \frac{3}{2} = 1.5 \)
\( \frac{a_5}{a_4} = \frac{5}{3} = 1.67 \)

Question. Let there be an A.P. with first term 'a', common difference 'd'. If \( a_n \) denotes its nth term and \( S_n \) the sum of first n terms, find. n and \( S_n \), if a = 5, d = 3 and \( a_n = 50 \). 
Answer: Given,
First term(a) = 5
Common difference(d) = 3
and, nth term (\( a_n \)) = 50
\( a + (n - 1)d = 50 \)
\( 5 + (n - 1)(3) = 50 \)
\( 5 + 3n - 3 = 50 \)
\( 3n = 50 - 5 + 3 \)
\( 3n = 48 \)

\( \implies \) \( n = \frac{48}{3} = 16 \)
Therefore, \( S_n = \frac{n}{2} [a + a_n] \)
= \( \frac{16}{2} [5 + 50] \)
= \( 8 \times 55 = 440 \)