# Arithmetic Progressions Examples and Solutions

## Notes

Ex. (1) A mixer manufacturing company manufactured 600 mixers in 3rd year and in 7th year they manufactured 700 mixers. If every year there is same growth in the production of mixers then find (i) Production in the first year (ii) Production in 10th year (iii) Total production in first seven years.

Solution : Addition in the number of mixers manufactured by the company per year is constant therefore the number of production in successive years is in A.P.
(i) Let’s assume that company manufactured tn mixers in the nth year then as per
given information,
t3 = 600, t7 = 700
We know that tn= a+(n-1)d
t3= a+(3-1)d
a + 2d = 600. . . (I)
t7= a+(7-1)d
t7= a+6d = 700
a+2d = 600

∴ Substituting a = 600 - 2d in equation (II),
600 - 2d + 6d = 700
4d = 100

∴ d = 25
a+2d = 600

∴ a + 2 ×25 = 600
a + 50 = 600

∴ a = 550
∴ Production in first year was 550.

(ii) tn= a+(n-1)d
t10= 550+(10-1) × 25
= 550 + 225
Production in 10th year was 775.

(iii) For finding total production in first 7 years let’s use formula for Sn.
S_n =n/2 [2a+(n-1)d]
S_n =7/2 [1100 + 150] =7/2 [1250] = 7 xx 625 = 4375
Total production in first 7 years is 4375 mixers.

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