Question

Find the sum of all the integers between 58 and 576 which when divided by 16 leave remainder 7

Answer 1

a=71
d=16
an=567
an=a+(n-1)d
567=71+(n-1)16
n-1=31
n=32
Sn=n/2 [2a+(n-1)d]
S32 = 32/2[2*71+ (32-1)16]
S32 = 16 * 638
S32=10208

Answer 2

Answer:

The sum of all the integers between 58 and 576 which when divided by 16 leave remainder 7 is 10208

Step-by-step explanation:

According to question,

Such numbers are 71, 87, 103, . . . . . . . , 551 , 567

We can see the common difference between all the numbers in series is same, that is, the given series is in AP

We know the $n^{th}$ term in AP is given as :

$T_{n}$ = a + ( n - 1 ) d

where,

$T_{n}$ = $n^{th}$ term

a = First term in series

n = Number of terms

d = Common difference

Substituting all the values,

567 = 71 + ( n - 1 ) *16

567 = 71 + 16n - 16

567 = 55 + 16n

16n = 567 - 55

16n = 512

n = 32

So,

The number of terms in this series is 32

Now, We need to find the sum of this series

We know,

The sum of N terms in AP with first and last term is given as :

$S_{n}$ = $\frac{n}{2}$ [ a + l ]

where,

Sn = Sum of series

n = Number of terms in series

a =  First term of series

l = Last term of series

Substituting the values ,

$S_{n}$ = $\frac{32}{2}$ [ 71 + 567 ]

$S_{n}$ = 16* [ 638 ]

$S_{n}$ = 10208

Hence,

The sum of all the integers between 58 and 576 which when divided by 16 leave remainder 7 is 10208