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The number of terms of an AP is even; the sum of the odd terms is 24, of the even terms is 30, and the last term exceeds the first term by 10.5. Find number of terms and series... The number of terms of an AP is even; the sum of the odd terms is 24, of the even terms is 30, and the last term exceeds the first term by 10.5. Find number of terms and series...
let the total no.of terms be 2n , & first term ,a & common diff. be d a + (2n-1 ) d = l & l-a = 10.5 if we take only the odd terms , it will also be a A.p. of n terms with common diff. of 2d 24 = n/2 { 2a + (n-1)2d } 24 = n { a + (n-1 )d } .................................(2) similarly, 30 = n/2 { 2(a+d) + (n-1)2d } 30 = n/2 { 2a + 2nd } 30 = n { a + nd } ...............................(3) putting the value of n { a + nd } from eq 3 , in eq 2, 24 = 30 - nd nd = 6 , putting in eq 1 12 - d = 10.5 d = 1.5 , so n = 6/ 1.5 = 4 putting in eq 3, 30 = 4 ( a + 6 ) a = 1.5 so the no. of terms = 2n = 8 & series , 1.5 , 3, 4.5 , 6, 7.5, 9, 10.5 . 12 (2n-1 ) d = 10.5 .............. ...............(1)
let the total no.of terms be 2n , & first term ,a & common diff. be d
a + (2n-1 ) d = l
& l-a = 10.5
if we take only the odd terms , it will also be a A.p. of n terms with common diff. of 2d
24 = n/2 { 2a + (n-1)2d }
24 = n { a + (n-1 )d } .................................(2)
similarly,
30 = n/2 { 2(a+d) + (n-1)2d }
30 = n/2 { 2a + 2nd }
30 = n { a + nd } ...............................(3)
putting the value of n { a + nd } from eq 3 , in eq 2,
24 = 30 - nd
nd = 6 , putting in eq 1
12 - d = 10.5
d = 1.5 , so n = 6/ 1.5 = 4
putting in eq 3,
30 = 4 ( a + 6 )
a = 1.5
so the no. of terms = 2n = 8
& series ,
1.5 , 3, 4.5 , 6, 7.5, 9, 10.5 . 12
(2n-1 ) d = 10.5 .............. ...............(1)
Dear student,Please find the solution to your problem below. Let the total no.of terms be 2n , & first term ,a & common diff. be da + (2n – 1)d = l & l – a = 10.5(2n – 1)d = 10.5 → (1) If we take only the odd terms, it will also be an A.P. of n terms with common diff. of 2d24 = n/2 {2a + (n – 1)2d}24 = n { a + (n-1 )d } → (2) Similarly,30 = n/2 {2(a+d) + (n – 1)2d}30 = n/2{2a + 2nd}30 = n { a + nd } → (3)putting the value of n { a + nd } from eq 3 , in eq 2,24 = 30 – ndnd = 6 , putting in eq 112 – d = 10.5d = 1.5 , so n = 6/ 1.5 = 4putting in eq 3,30 = 4 ( a + 6 )a = 1.5 so the no. of terms = 2n = 8 & series is :- 1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12 Hope it helps.Thanks and regards,Kushagra
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