Question

11,15,19,23 find the sum of all terms which are less than 100

Answer 1

Step-by-step explanation:

The two series are in A.P. Therefore, the common series will also be in an A.P

Common difference of 1st series =4 and the common difference of 2nd series =5.

Common difference of the sequence whose terms are common to the two series is given by L.C.M of 4 and 5 =20

Here the first term of the identical (common terms) sequence is 19.

We know the sum of first n terms is given by =

2

n

[2a+(n+1)d]

Hence, the sum of first 50 terms of this sequence =

2

50

[2×19+(50+1)20]=25450.

Hope it helps....

Answer 2

$\sf\huge\bold{Answer:}$

Given :

Series : 11,15,19,23,...

To find :

Sum of all terms in series less than 100

Solution :

To prove that this series is an AP

we have to prove that d1 = d2

d1 = a2-a1 = 15-11 = 4

d2 = a3-a2 = 19-15 = 4

Hence, d1=d2

AP formed is

11,15,19,23,...

a5 = a4+d = 23+4=27

a6=a5+d= 27+4 = 31

a7=a6+d=31+4=35

a8=a7+d=35+4=39

As we can see that the coming up terms as well as given terms have 1,5,9,3,7 on ones place i.e. all odd numbers.

Hence the closest odd number to 100 that is less than 100 is 99.

We got a(n)= 99

AP is 11,15,19,23,..., 99

Formula to find number of terms in AP (n)

a(n)=a+(n-1)d

99=11+(n-1)4

99-11=4n-4

88+4 =4n

84=4n

84/4=n

n=21

There are 21 terms in this AP

Now, sum of all terms

Formula used :

S(n)=n/2[2a+(n-1)d]

S(n) = 21/2 [2(11)+(21-1)4]

S(n)=21/2[22+20(4)]

S(n)=21/2[22+80]

S(n)=[21×102]/2

S(n)=21×51

S(n)=1071

The sum of all terms which are less than 100 is 1071.