Solved Word Problems in Arithmetic Progression :
Here we are going to see some practice questions on arithmetic sequence.
Question 1 :
In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.
Let the temperature of Ooty from Monday to Friday are
a - 2d, a - d, a, a + d , a + 2d
a - 2d + a - d + a = 0
3a - 3d = 0
a - d = 0 ---(1)
a + a + d + a + 2d = 18
3a + 3d = 18
a + d = 6 ---(2)
(1) + (2)
a + a = 6
2a = 6
a = 3
By applying the value of a in (1), we get
3 + d = 6
d = 6 - 3
d = 3
a - 2d = 3 - 2(3) = -3° C
a - d = 3 - 3 = 0° C
a = 3° C
a + d = 3 + 3 = 6° C
a + 2d = 3 + 2(3) = 9° C
Question 2 :
Priya earned 15,000 in the first month. Thereafter her salary increased by 1500 per year. Her expenses are 13,000 during the first year and the expenses increases by 900 per year. How long will it take for her to save 20,000 per month.
Increment of salary per year = 1500
monthly increment = 1500/12 = 125
write the expenses as a sequence
1st month expense = 13000
increasing expenses per year = 900
monthly increment of expenses = 900/12 = 75
Let us write the monthly earnings and expenses as a sequence.
15000, 15125, 15250,................
13000, 13075, 13150,...........
By subtract these two sequences, we get saving of each month
2000, 2050, 2100,................
From the above sequence, we may find that saving is increased 50 per month.
To get in which month will she save 20000, we have to use the formula
n = [(l - a)/d] + 1
= [(20000 - 2000)/50] + 1
n = 361 (number of months)
In 361 months, she will save 20000 per month. But we have to find the answer in years.
= 361/12 = 30.8
Hence the required number of years is 31.
After having gone through the stuff given above, we hope that the students would have understood, "Word Problems in Arithmetic Sequence".
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