ncert solutions for class 7 maths

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2. https://www.entrancei.com/ ARITHMETIC PROGRESSION 1. Arithmetic Progression (A.P.). Arithmetic Progression (or Arithmetic sequence) is a sequence in which the difference of a term and its predecessor is always constant, i.e., common difference. The successive ......... 4 , 3 , 2 , , d a d a d a d a a + + + + . − = where d is the A.P. a a of d an , − n n 1 terms are • Finite A.P.: An A.P. containing finite number of terms is called finite A.P. e.g. 147, 149, 151 ………………….. 163. • Infinite A.P.: An A.P. containing infinite terms is called infinite A.P. e.g. 6, 9, 12, 15 ……………………….. For Example: Which of the following are A.P’s? If they form an A.P., find the common difference d and the next three terms after last given term. 12, 52, 72, 73, … (i) ... (ii) , 3 , 6 , 9 (i) , 3 , 6 , 9 ... − = − = ) 1 − a a 6 3 3 ( 2 2 1 − = − = − a a 9 6 3 ( 3 2 ) 3 2 .......... .......... .......... .......... .......... .......... .. .......... .......... .......... .......... .......... .......... .. As the common difference between any two consecutive terms is not the same  The given sequence is not an A.P. 2 2 2 1 5 , 7 , 73 , ,.... (ii) 2 − = − = − = a a 5 1 25 = 1 − 24 2 1 2 2 − = − = a a 7 5 49 25 24 3 2 −a = − = a 73 49 24 4 3 .......... .......... .......... .......... ......... .......... .......... .......... .......... ......... As the common difference between any two consecutive terms is the same  The given sequence is an A.P. Next three terms are: 121 24 97 97 24 73 6 5 = + = = + = a a 145 24 121 7 = + = a 2. General Term. General term or nth term of an A.P. is = + ) 1 − where a = first term, d = an a ( n d , https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/ common difference. • • The formula quantities being given, the fourth can be found by using above relation. = + ( − contains four quantities n a , a, n and d. Three an a n ) 1 d For Example: Find the 18th term and nth term for the sequence 7, 4, 1, −2, −5. Here a = 7 and d = 1 2 a a − = 4 − 7 = − 3 n = 18 ( )d n a an 1 − + = ( ) 3 1 18 7 18 −  − + = a = 7 + 17 −3 = 7 − 51 = −44 ( )d n a an 1 − + = = 7 + (n− 1) (−3) = 7 − 3n + 3 = 10 − 3n • • If only two quantities are given, two conditions (equations) in the problem should be given. Therefore, to determine these two unknowns, we have to solve both the conditions (equations) linearly. For Example: The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms. 3 a = 7 7 a = 2 33+ a a + 2d = 7 ..... (i) a + 6d = 3 × 7 + 2 a + 6d = 23 ..... (ii) Solving (i) and (ii) 4d = 16 d = 4 Putting the value of d in equation (i), a + 2 × 4 = 7 a = – 1 20  − + −   [( 2 ( ) 1 ( 20 ) 1 ] 4 = S 20 2 3. So, = 10 [–2 + 76] = 10 (74) = 740 d = 4 a = –1 S S 20 20 First negative term To find the first negative term of the given sequence, put  . a 0 n For example: https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ 1 1 3 To find the first negative term of the sequence 20, . 19 18 , 17 , ,.... 4 2 4 3 = = − Here a 20 and d 4  ( Then,  a a 0 n + ) 1 −  n d 0 −  3  + n ) 1 −   20 ( 0     4 83 3 n  −  0 4 4  − n  83 3 0 2       3 n 83 n 27 n 28 3 Thus, 28th term of the given sequence is the first negative term. Finding term from the end To find the nth term from the end, apply l is the last term. or take last term (l) of the given sequence as ‘a’ and take common difference as (−d) For example: Find the 5th term from the end of the AP, 17, 14, 11, ….., −40 1st method 3 17 14 , 40 − = − = − = d l 4. = + − − = − ( − or , where an l ( n 1 )( d ) an l n ) 1 d − ( − Using n ) 1 d l 5th term from the end will be ) 1 5 ( 40  − − − = − = 40 − −  − 3 4 3 = −28 = 40 + − 12 2nd method Sequence can be written as −40, −37, …. 11, 14, 17  = − a 40 = −37 + 40 = 3 = 37 − − 40 − d ( ) n = 5 = + ( − Using an a n ) 1 d − + −  = 40 − +  = 40 ( 5 ) 1 3 4 3 = − 40 + 12 = −28 5. Condition for terms to be in A.P. If three terms a, b, c are in A.P. then − = − = + i.e., b a c , b 2 b a c . For example: If , 2 + x x + are in A.P., find the value of x. 10 3 , x 2 + + Since, are in A.P. 2 x , x 10 3 , x 2 https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ ( ) ( )  + = + + 2 x 10 2 x 3 x 2  + = + 2 x 20 5 x 2  3 = x 18  6. x = 6 Choice of terms in A.P. No. of terms Terms Common difference − + a d , a , a d 3 d − − + + a 3 d , a d , a d , a 3 d 4 2d − − + + a 2 d , a d , a , a d , a 2 d 5 d − + − + − + a a 5 3 d d , , a a 3 5 d d , a d , a d , 6 2d For example: The sum of three numbers in A.P. is -3, and their product is 8. Find the numbers. Let the number be ( ) ( ) d a a d a + − , , . Then ( ) ( ) 3 3 − = + + + −  − = d a a d a Sum  = − 3 a 3  = − …(i) a 1 Now, Product = 8 ( a ( a a ( − )( )( a )  − + = d a ) d 8 2 2  −d = ) 8 )( 2=  − [from (i)] 1 1 d 8 2  =  =  d 9 d 3 , 4− − If d = 3 the numbers are If d = -3, the numbers are 2, -1, -4. , 1 . 2 , 4− − , 1 − . 4 − 7. Thus, the numbers are Sum of first n-terms of an A.P. or , 1 2 , 2 n = + ) 1 − Sn 2 [ a ( n d ] 2 8. where a = first term, d = common difference. Sn is also given by the expression n = + Sn ( a l ), 2 = + ) 1 − = where l = last term as l a ( n d a . n For example: The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. https://www.entrancei.com/ncert-solutions-class-10-maths

6. https://www.entrancei.com/ = = = = Here, a , 5 a 45 , S 400 l n n n 2   l = +  Sn a n 2    = + = 400 5 45 25 n 400=  = n 16 25 Also l = 45 ( )  + ) 1 − = an = a n 1 d 45 (  + −  = = = 5 16 d 45 [ a , 5 n 16 ]  = − = 15 d 45 5 40 40= 8  = d 15 3 8 = = Hence, number of terms(n) 16 , d 3 9. of n terms of a sequence is given then nth term = n a If sum can be determined by nth term of the sequence S (n a ( ) ) n − and common difference S S − n n 1 = − − −+ = . d a a S 2 S S − − n n 1 n n 1 n 2 For example: 2 n − 3 2 n , find the A.P. and its 19th term. If the sum of first n terms of an A.P. is 2− = Sn 3 n 2 n = − = Put n = 1 ; S 3 2 1 1 2 = − Put n = 2 ; S ( 3 ) 2 ( 2 ) 2 2 = 3 × 4 – 4  8 2 = − Put n = 3 ; S ( 3 3 ) ( 2 3 ) 3 = 27 – 6 = 21  =S 1= a 1 = − = a S S 7 2 2 1 = − = − = a S S 21 8 13 3 3 2  Required A.P. is; 1, 7, 13, .......... a = 1, d = 6 = + a a 18 d 19 = 1 + 18 × 6 = 1 + 108 = 109 https://www.entrancei.com/ncert-solutions-class-10-maths

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