Class 10 math exercise 5.2 question 7
Ncert class 10 math exercise 5.2 question 7
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution
Let a and d is first term and common difference of the AP.
According to Question,
11th term is 38
means, a11=38
a+10d=38 ............(i)
and 16th term is 73.
means, a16=73
a+15d=73 .............(ii)
From Equ (i) & (ii)
a+10d=38 ............(i)
a+15d=73 .............(ii)
After solving,
-5d=-35
5d=35
d=7
value of d put in equation (i)
a+10×7 =38
a= 38-70
a=-32
now the 31st term of an AP
means, a31=a + 30d
a31=-32 + 30×7
a31=-32 +210
a31= 178
Hence the 31st term of an AP=178
Q.The 15th term of an arithmetic series is 143 and 31st term is 183.
- Find the first term and common difference.
- Find the 100th term of the series.
Solution
Let a and d is first term and common difference of the AP.
According to Question,
15th term of an arithmetic series is 143
means, a15 =143
a+14d =143 ..............(i)
and 31st term is 183.
means, a31 =183
a+30d =183 ..............(ii)
From equation (i) & (ii)
a+14d =143 ..............(i)
a+30d =183 ..............(ii)
After solving,
a= 108
d=5/2
1.first term & common difference
a= 108
d=5/2
2.the 100th term of the series
Means, a100 =a+99d
a100 =108+99×5/2
a100 =108+495/2
a100 =(216+495)/2
a100 =711/2 Answer...
Q.If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is −3 then find the 10th term.
Let a and d is first term and common difference of the AP.According to Question,
sum of 3rd and 8th terms of an A.P. is 7
means, a3 + a8 =7
a + 2d + a + 7d = 7
2a + 9d =7 ..............(i)
sum of 7th and 14th terms is −3
means, a7 + a14 =-3
a + 6d + a +13d =-3
2a + 19d =-3 ..............(ii)
From Equation (i) & (ii)
2a + 9d =7 ..............(i)
2a + 19d =-3 ..............(ii)
After solving,
-10d=10
d=-1
value of d put in Equation (i)
2a + 9×(-1)=7
2a -9=7
2a=16
a=8
the 10th term=
means,
a10 = a + 9d
a10 = 8 + 9×(-1)
a10 = 8 - 9
a10 = -1
hence the 10th term= -1
