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Grade: 11

                        

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...

11 years ago

Answers : (2)

Pratham Ashish
17 Points
							

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)

 

11 years ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
Dear student,
Please find the solution to your problem below.
 
Let the total no.of terms be 2n , & first term ,a & common diff. be d
a + (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 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 is :- 1.5, 3, 4.5, 6, 7.5, 9, 10.5, 12
 
Hope it helps.
Thanks and regards,
Kushagra
3 months ago
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