ARITHMETIC PROGRSSION AND GEOMETRIC PROGRESSION - SOLUTIONS 1 b) Given that, q d p a T p = − + = ) 1 ( .....(i) and p d q a T q = − + = ) 1 ( ..... (ii) From (i) and (ii), we get 1 ) ( ) ( − = − − − = q p q p d Putting value of d in equation (i), then 1 − + = q p a Now, th r term is given by A.P. ) 1 )( 1 ( ) 1 ( ) 1 ( − − + − + = − + = r q p d r a T r r q p − + = Note : Students should remember this question as a formula. 2. (b) The sum of integers from 1 to 100 that are divisible by 2 or 5 = sum of series divisible by 2 + sum of series divisible by 5 – sum of series divisible by 2 and 5. ) 100 ....... 15 10 5 ( ) 100 ...... 6 4 2 ( + + + + + + + + = ) 100 ........ 30 20 10 ( + + + + − 5 ) 1 20 ( 5 2 2 20 2 ) 1 50 ( 2 2 2 50 − + + − + = 10 ) 1 10 ( 2 10 2 10 − + − 3050 550 1050 2550 = − + = 3. (c) 1 3 , 8 , 2 + + x x x are in A.P. Therefore 2 1 5 2 ) 1 3 ( ) 2 ( ) 8 ( + = + + = + x x x x 1 5 16 2 + = + x x 15 3 = x 5 = x 4. (b) 35 8 9 = + = d a T and 75 18 19 = + = d a T Solving the equations, we get 4 = d and 3 = a Hence th 20 term of A.P. 79 4 19 3 19 = + = + d a 5. (a) Required ratio is 9 4 99 44 = 6. (b) Obviously, 1 1 2 1 2 3 = − = − = a a 0 1 1 1 3 4 = − = − = a a 1 1 0 1 4 5 − = − = − = a a 7. (c) Let D be the common difference of the A.P. Then, = + − + − e d c b a 4 6 4 0 ) 4 ( ) 3 ( 4 ) 2 ( 6 ) ( 4 = + + + − + + + − D a D a D a D a a 8. (a) } ) 1 ( { } ) 1 ( { d q a q d p a p − + = − + 0 ) ( ) ( ) ( 2 2 = − + − + − d p q d q p q p a 0 } ) 1 ( ){ ( = − + + − d q p a q p 0 ) 1 ( = − + + d q p a 0 = + q p T , } { q p 9. (a) We have 5 6 3 2 2 1 + + = n n S S n n 5 6 3 2 ] ) 1 ( 2 [ 2 ] ) 1 ( 2 [ 2 2 2 1 1 + + = − + − + n n d n a n d n a n 5 6 3 2 2 1 2 2 1 2 2 2 1 1 + + = − + − + n n d n a d n a 5 6 3 2 2 1 2 1 2 2 1 1 + + = − + − + n n d n a d n a Put 25 = n then 3 ) 25 ( 6 3 ) 25 ( 2 12 12 2 2 1 1 + + = + + d a d a 155 53 2 1 13 13 = T T 10. (d) It is not possible to express e d c b a + − + + 4 4 in terms of a 11. (d) Given series ........ 6 1 3 1 2 1 + + + Here 2 1 = a , common difference 6 1 − = d and 9 = n 2 3 6 1 ) 1 9 ( 2 1 2 2 9 9 − = − − + = S 12. (c) Since q d p a T p 1 ) 1 ( = − + = .....(i) ......(i) and p d q a T q 1 ) 1 ( = − + = .....(ii) From (i) and (ii), we get pq a 1 = and pq d 1 = Now sum of pq terms − + = pq pq pq pq 1 ) 1 ( 2 2 2 1 2 1 2 ) 1 ( 2 1 1 2 2 + = − + = − + = pq pq pq pq pq Note : Students should remember this question as a formula. 13. (d) We know that natural numbers are n ........ , 4 , 3 , 2 , 1 are in A.P. Now sum ) 1 ( 2 ] 1 ) 1 ( 1 2 [ 2 + = − + = n n n n 14. (a) First term 2 = a and common difference 4 = d and 40 = n Then ] ) 1 ( 2 [ 2 d n a n S n − + = ] 156 4 [ 20 ] 4 39 4 [ 20 + = + = 3200 20 160 = = 15. (b) Series, ....... 11 8 5 2 + + + + 3 , 2 = = d a and let number of terms is n then sum of A.P. } ) 1 ( 2 { 2 d n a n − + = 3 ) 1 ( 2 2 2 60100 − + = n n ) 1 3 ( 120200 + = n n 0 120200 3 2 = − + n n 0 ) 601 3 )( 200 ( = + − n n Hence 200 = n 16. (b) We get series 3, 6, 9, 12, ........ 99. Here 3 , 3 , 33 3 99 = = = = d a n , therefore 3 ) 1 33 ( 3 2 2 33 − + = S 1683 51 33 102 2 33 = = = 17. (c) .......... 7 5 3 1 + + + + upto n terms. 2 2 ) 1 ( 1 2 2 n n n S n = − + = 18. (a) According to condition ) 3 )( 1 ( 54 2 2 513 − − + = n n ) 3 111 ( 1026 n n − = 0 1026 111 3 2 = + − n n 0 ) 18 )( 57 3 ( = − − n n Hence 18 = n 19. (a) Given that n n S n 5 2 2 + = Putting 7 1 5 1 2 , .......... , 3 , 2 , 1 1 = + = = S n , 33 15 18 , 18 10 8 10 4 2 3 2 = + = = + = + = S S So, 11 7 18 , 7 1 2 2 1 1 = − = − = = = = S S T a S T , 15 18 33 2 3 3 = − = − = S S T Therefore series is ........ , 15 , 11 , 7 Now, th n term 3 4 4 ) 1 ( 7 ) 1 ( + = − + = − + = n n d n a Aliter : As we know 1 − − = n n n S S T ) 1 ( 5 ) 1 ( 2 ) 5 2 ( 2 2 − + − − + = n n n n 3 4 5 5 2 4 2 5 2 2 2 + = + − − + − + = n n n n n n 20. (d) Here 1 3 − = n T n , putting 5 , 4 , 3 , 2 , 1 = n we get first five terms, 14 , 11 , 8 , 5 , 2 Hence sum is 40 14 11 8 5 2 = + + + + Aliter : n n n n T S n n − + = − = = 2 ) 1 ( 3 1 3 40 5 2 6 5 3 5 = − = S 21. (c) Given that first term 10 = a , last term 50 = l and sum 300 = S ) ( 2 l a n S + = ) 50 10 ( 2 300 + = n 10 = n 22. (a) Series 999 ........ 117 108 + + + is an A.P. where 108 = a , common difference 9 = d , 100 11 111 9 99 9 999 = − = − = n Hence required sum = 55350 1107 50 ) 999 108 ( 2 100 = = + 23. (c) Given that 2 2 ] ) 1 ( 2 [ 2 ] ) 1 ( 2 [ 2 n m d n a n d m a m = − + − + n m d n a d m a = − + − + ) 1 ( 2 ) 1 ( 2 n m d n a d m a = − + − + ) 1 ( 2 1 ) 1 ( 2 1 md n am nd m an ) 1 ( 2 1 ) 1 ( 2 1 − + = − + 0 ] [ 2 ) ( = + − − + − m mn n mn d m n a 0 ) ( 2 ) ( = − + − n m d m n a 2 d a = or a d 2 = So, required ratio, a n a a m a d n a d m a T T n m 2 ) 1 ( 2 ) 1 ( ) 1 ( ) 1 ( − + − + = − + − + = 1 2 1 2 2 2 1 2 2 1 − − = − + − + = n m n m Trick : Replace m by 1 2 − m and n by 1 2 − n . Obviously if m S is of degree 2, then m T is of 1 e i linear. 24. (d) Let 100 .......... 3 2 1 + + + + = S 5050 ) 101 ( 50 ) 100 1 ( 2 100 = = + = Let 99 ......... 12 9 6 3 1 + + + + + = S = ) 33 ......... 4 3 2 1 ( 3 + + + + + = 1683 17 99 ) 33 1 ( 2 33 3 = = + Let 100 ........ 15 10 5 2 + + + + = S = ) 20 ........ 3 2 1 ( 5 + + + + = 1050 21 50 ) 20 1 ( 2 20 5 = = + Let 90 ........ 45 30 15 3 + + + + = S = ) 6 ........ 3 2 1 ( 15 + + + + = 315 7 45 ) 6 1 ( 2 6 15 = = + Required sum = 3 2 1 S S S S + − − = 315 1050 1683 5050 + − − = 2632. 25. (c) Let first 3 terms be a d a , − and d a + Now 12 ) ( ) ( = + + − d a d a 12 2 = a 6 = a and 24 ) ( = − a d a 24 ) 6 ( 6 = − d 2 = d First term = 4 2 6 = − = − d a 26. (d) The number divisible by 3 between 250 to 1000 are 252, 255, .........., 999. 3 ) 1 ( 252 999 − + = = n T n 1 84 333 − + = n 250 = n ] 999 252 [ 2 250 ] [ 2 + = + = l a n S 27. (b) th 7 term of an A.P. 40 = 40 6 = + d a )] 6 ( 2 [ 2 13 ] ) 1 13 ( 2 [ 2 13 13 d a d a S + = − + = 520 40 2 2 13 = = 28. (b) Obviously )} 1 ( 5 ) 1 ( 3 { ) 5 3 ( 164 2 2 − + − − + = m m m m 5 5 3 6 3 ) 5 3 ( 2 2 + − − + − + = m m m m m 2 6 164 + = m 27 = m 29. (b) 56 ) 3 11 ( ) 2 11 ( ) 11 ( 11 = + + + + + + d d d 2 = d and } 2 ) 2 ( 11 { } 2 ) 1 ( 11 { − + + − + n n 112 } 2 ) 4 ( 11 { } 2 ) 3 ( 11 { = − + + − + + n n 11 = n 30. (d) ] ) 1 ( 2 [ 2 d n a n S − + = 4 ) 1 ( 6 2 406 − + = n n ] 4 4 6 [ 812 − + = n n 2 4 2 812 n n + = n n + = 2 2 406 0 406 2 2 = − + n n 2 2 406 2 4 1 1 + − = n 4 3249 1 − = 4 57 1 − = Taking (+) sign, 14 4 57 1 = + − = n 31. (a) Under conditions, we get − + = − + d a d a ) 1 5 ( 2 2 5 4 ) 1 10 ( 2 2 10 d a d a 8 4 9 2 + = + or 2 1 = d a Hence 2 : 1 : = d a 32. (a) We have 7 terms 10 ...upto ..... 11 8 5 terms upto ...... 7 5 3 = + + + + + + n 7 37 10 ) 4 2 ( 7 ] 3 ) 1 10 ( 10 [ 2 10 ] 2 ) 1 ( 6 [ 2 = + = − + − + n n n n 0 ) 35 )( 37 ( 0 1295 2 2 = − + = − + n n n n Hence 35 = n 33. (c) Let the sides of the triangle be d a a d a + − , , , then hypotenuse being the greatest side e i d a + So, 2 2 2 ) ( ) ( d a a d a − + = + d a d ad a a ad d a 4 2 2 2 2 2 2 2 = + − + = + + Therefore ratio of the side d a a d a s + − = : : 5 : 4 : 3 ) 4 ( : 4 : ) 4 ( = + − = d d d d d 34. (b) Since q r q p + + 1 , 1 and r q + 1 are in A.P. p r r q q p q r + − + = + − + 1 1 1 1 ) )( ( ) )( ( p r r q r q p r q p p r p r q p + + − − + = + + − − + r q q p q p r q + − = + − or 2 2 2 2 q p r q − = − 2 2 2 2 p r q + = Therefore 2 2 2 , , r q p are in A.P. 35. (c) Let the first term of A.P. is a and common difference is d 11 th term of A.P. = d a 10 + 21 st term of A.P. = a + 20 d ) 20 ( 7 ) 10 ( 2 d a d a + = + d a d a 140 7 20 2 + = + 0 120 5 = + d a 0 24 = + d a Hence 25 th term is 0.