Suppose $$a_1, a_2, \ldots, a_n$$ be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms to the sum of first nine terms of the progression is $$5:12$$and $$110 < a_{15} < 120$$, then the sum of the first ten terms of the progression is equal to 

Solution :

Let the arithmetic progression be :

$$a,\ a+d,\ a+2d,\ldots$$

Sum of first $$n$$ terms of an AP :

$$S_n = \frac{n}{2}[2a+(n-1)d]$$

Given :

$$\frac{S_5}{S_9} = \frac{5}{12}$$

Now,

$$S_5 = \frac{5}{2}[2a+4d]$$

$$= 5(a+2d)$$

and,

$$S_9 = \frac{9}{2}[2a+8d]$$

$$= 9(a+4d)$$

Therefore,

$$\frac{5(a+2d)}{9(a+4d)}=\frac{5}{12}$$

Cancelling 5 :

$$\frac{a+2d}{9(a+4d)}=\frac{1}{12}$$

$$12(a+2d)=9(a+4d)$$

$$12a+24d=9a+36d$$

$$3a=12d$$

$$a=4d$$

Now,

$$a_{15}=a+14d$$

$$=4d+14d$$

$$=18d$$

Given :

$$110<a_{15}<120$$

$$110<18d<120$$

$$\frac{110}{18}<d<\frac{120}{18}$$

$$6.11<d<6.67$$

Since $$d$$ is a natural number,

$$d=6$$

Hence,

$$a=4d=24$$

Now sum of first 10 terms :

$$S_{10}=\frac{10}{2}[2a+9d]$$

$$=5[2(24)+9(6)]$$

$$=5[48+54]$$

$$=5(102)$$

$$=510$$

Final Answer :

$$510$$