Let us assume the first term is a and common difference is d.
Formula for nth term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t1 = a + ( n - 1 ) x d
t1 = a + (1 - 1) x d = a + 0 x d = a
Fifth term t5 = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t1 + t5 = 10
? a + a + 4d = 10
? 2a + 4d = 10
? a + 2d = 5.............................(1)
Formula of sum of n terms for Arithmetic Progression,
Sn = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
? S10 = 10/2[ 2a + ( 10 - 1 ) x d ]
? S10 = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S10 - first term = 99
? 10a + 45d - a = 99
? 9a +45d = 99
? a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
? a + 5d - a - 2d = 11 - 5
? 3d = 6
? d = 2
Put the value of d in equation (1), we will get
? a + 2 x 2 = 5
? a + 4 = 5
? a = 5 - 4
? a = 1
Since tn = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t3 = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.
Here sum is put on compound interest,
? P.W. = A / (1 + r / 100)n = 2420 / (1 + 10 / 100)2 = 2420 x 100 / 121 = Rs. 2000
? T.D. = P.W. - P
? True discount = 2420 - 2000 = Rs. 420
The Sum of the digits in each number, Except 324 is 10.
The given number series follows the pattern that,
24×0 + 4 = 4
4×1 + 9 = 13
13×2 + 16 = 42
42×3 + 25 = 151
151×4 + 36 = 640
Therefore, the odd number in the given series is 41
From the beginning, the next term comes by adding prime numbers in a sequence of 2, 3, 5, 7, 9, 11, 13... to its previous term. But 165 will not be in the series as it must be replaced by 166 since 153+13 = 166.
The given number series follows a pattern that
196, 169, 144, 121, 100, 81, ?
-27 -25 -23 -21 -19 -17
=> 81 - 17 = 64
Therefore, the series is 196, 169, 144, 121, 100, 81, 64.
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