Test: Word Problems of AP - Class 10 MCQ

We know S1is the sum of 1st term which is a
S₁ = 3 + 5 = 8 = a
S₂ is sum of first two terms
S₂=3 * 4 + 5 * 2 = 12 + 10 = 22  
And We have a₂= S₂ - S₁ = 22 - 8 = 14
d = a₂ -a₁ = 14 - 8 = 6 

sum=n/2x(2a+(n-1)d )
       =50/2(2x3+ (50-1)3 )
       = 25 ( 6+49x3 )
       = 25x 153
       = 3825

Distance increased every day is the common difference for the AP which is 0.5
Initial distance is 5 km , so a=5
To find the distance at 10th day, so n=10
l=a+(n-1)d
 =5+(10-1)0.5
 =5+4.5=9.5km

One metre = 100 cm, a = 100 cm

since every year it grows 4 cm less than it did the preceding year, d = -4

Thus, it is an AP, with initial term a = 100 and d = -4

The tree will stop its growth, when nth term of the series is 0.

Nth term = a + (n - 1)d

0 = 100 + (n - 1)(-4)

0 = 100 - 4n + 4

0 = 104 -4n

4n = 104

n = 104/4

n = 26 years