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Question

Mathematics Question on Arithmetic Progression

The number of terms in an A.PA.P. is even; the sum of the odd terms in it is 2424 and that the even terms is 3030. If the last term exceeds the first term by 1012,10 \frac{1}{2}, then the number of terms in the A.PA.P. is :

A

4

B

8

C

12

D

16

Answer

8

Explanation

Solution

Let no. of terms =2n=2n
a,(a+d),(a+2d)a ,( a + d ),( a +2 d ), ,........ a+(2n1)da +(2 n -1) d
sum of even terms
n2[2(a+d)+(n1)2d]=30......(i)\frac{ n }{2}[2( a + d )+( n -1) 2 d ]=30 \,\,\,\,\, ......(i)
sum of odd terms
n2[2a+(n1)2d]=24.......(ii)\frac{ n }{2}[2 a +( n -1) 2 d ]=24\,\,\,\,\, .......(ii)
a+(2n1)da=212......(iii)a +(2 n -1) d - a =\frac{21}{2}\,\,\,\,\, ......(iii)
e (i)....e (ii)
n2×2d=6\frac{ n }{2} \times 2 d =6
nd=6......(iv)\Rightarrow nd =6 \,\,\,\, ......(iv)
(2n1)d=212......(v)(2 n -1) d =\frac{21}{2} \,\,\,\, ......(v)
eq(iv)eq(v)=n2n1=47\frac{ eq ( iv )}{ eq ( v )}=\frac{ n }{2 n -1}=\frac{4}{7}
8n4=7n\Rightarrow 8 n -4=7 n
n=4n =4
so no. of terms =8=8