The first number in each group: 1, 2, 5, 10, 17.....
Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.
Let the quadratic equation of the general term be
ax2+bx+c 1st term
=a+b+c=1 2nd term
=4a+2b+c=2 3rd term
=9a+3b+c=5 Solving the equations, we get
a=1,b=−2,c=2. Hence, the first term in the 15th group will be
(15)2−(15)2+2=197 We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in15th group will be 29. Hence, the last term in the 15th group will be
197+29−1=225. Sum of terms in group
15=(197+225)=6119 Alternatively,
The final term in each group is the square of the group number.
In the first group 1, second group 4, ............
The final element of the 14th group is
(14)2 , similarly for the 15th group this is :
(15)2=225 Each group contains all the consecutive elements in this range.
Hence the 15th group the elements are:
(197, 198, ................................225).
This is an Arithmetic Progression with a common difference of 1 and the number of element 29.
Hence the sum is given by :
(firstterm+lastterm) =×(197+225) =6119