The sum of first 10 terms of an arithmetic progression is 230.The sum of the first 20
terms is 860.What is the sum of the first 30 terms of the AP?.
Q 10 Which term of the arithmetic progression 11, 24, 37, 50 is 921?
Answers
SOLUTION - 1 :
GIVEN :
Sum of first 10 terms of an AP = 230
Sum of first 20 terms of an AP = 860
Sum of first 30 terms of an AP = ?
We know that,
Sum of terms in an AP = Sn = n/2 [ 2a(n - 1)d ]
S10 = 10/2 [ 2a + (9)d ]
230 = 5 [ 2a + 9d ]
230/5 = 2a + 9d
46 = 2a + 9d ------- (1)
S20 = 20/2 [ 2a + (19)d ]
860 = 10 [ 2a + 19d ]
86 = 2a + 19d ------ (2)
Solve eq - (1) and (2) to find (d)
2a + 9d = 46
2a + 19d = 86
(-)
---------------------
-10d = -40
d = 40/10
d = 4
Common Difference (d) = 4
Substitute d in eq - (1) to find first term(a)
2a + 9d = 46
2a + 9(4) = 46
2a + 36 = 46
2a = 46 - 36
a = 10/2
a = 5
First Term (a) = 5
Now,
Sum of first 30 terms,
= n/2 [ 2a + (n - 1)d]
= 30/2 [ 2(5) + (29)4]
= 15 [ 10 + 116 ]
= 15 [ 126 ]
= 1890
Therefore, sum of first 30 terms is 1890.
SOLUTION - 2 :
GIVEN :
AP = 24, 37, 50,......
First Term = a = 24
Common Difference = d = 37 - 24 = 13
nth term = 921
We know that,
In an AP, nth term = a + (n - 1)d
921 = 24 + (n - 1)13
921 = 24 + 13n - 13
921 - 24 = 13n - 13
897 = 13n - 13
897 + 13 = 13n
910 = 13n
n = 910/13
n = 70
Therefore, 921 is 70th term of an AP.
Answer:
The sum of first 10 terms is 230 then ,
n= 10 , S=230
S=n\2 [2a+(n-1)d]
230 =10\2[ 2a + (10 - 1)d ]
230= 5[ 2a+9d]
46= 2a+9d ........(1)
In second position ,
S=n\2 [2a+( n-1)d]
860 = 20\2 [2a+( 20-1)d]
860= 10[2a+19d]
86 = 2a +19d........(2)
after solving first and second
equations,
d=4 and a=5 ,
sum of first 30 terms of that AP = n\2 [2a+( n-1)d]
S = 30\2[ 2×5 +(30-1)4]
S= 15[ 10+116]
S= 15×126
S= 1890
So, the sum of 30 terms is 1890.