M4maths Previous Puzzles

31 Mar 2022 Arithmetic

Sum of first 13 terms of an arithmetic progression is 312. Which of the following can be the sum of the first 14 terms, if the first term and the common difference are positive integers?

OPtion

1) 356
2) 362
3) 350
4) 346
5) 370
6) 368
7) 359
8) 341
9) 355
10) None of these

Solution

In A.P. when First term=a, Difference=d, then Sum of n terms, Sn = (n/2)[2a + (n-1)*d]
.'. 312 = (13/2)*(2a + 12d)
48 = 2a + 12d
Now, Sum of first 14 terms = (14/2)*(2a + 13d) = 7(2a + 12d + d) = 7(48+d) = 336 + 7d
So, from the given options, only 350 is in the form of 336+7d

Correct Option: 3 Best Solution (2)

G.S.Kantharao   2 years AGO

S13=13(2a+12d)/2=312
13a+78d=312
a+6d=24
S14-S13=7(2a+13d)-(13a+78d)=14a+91d-13a+78d=P-312
a+13d=P-312
a+6d=24
7d=P-336
If P=350,then d=2 and a=12
Now S13=13(2*12+12*2)/2=24*13=312
S14=14(24+13*2)/2
=7*50=350
My option is 3👈

Like? Yes (4) | No   

Pargat Pal Singh   2 years AGO

(13/2)*(2a+12d)=312
a+6d=24
a=24-6d
As per given conditions, 'd' can have values 1,2,3.
Considering these values and given options, d=2, a=12 satisfies conditions.
Sum of first 14 terms = (14/2)*(2*12+13*2)=7*50=350

Like? Yes (3) | No (1)