Word problems based on APs

This is the decreasing AP (52,48,44,\ldots) and (S_{10}=340). Exam tip: treat the decrease as a negative common difference.

The plant numbers are (10,14,18,\ldots) and (S_{12}=384). Exam tip: take the total number of rows as (n).

The salary forms an AP and (a_8=12000+7(500)=15500). Exam tip: find the monthly salary using the (n)th term.

The practice times form the AP (25,30,35,\ldots) and (S_{14}=805). Exam tip: the same sum formula works when time increases regularly.

The donation amounts are (200,250,300,\ldots) and (S_{10}=4250). Exam tip: first and last day amounts are useful for total donation.

The matchstick numbers are (6,10,14,\ldots) and (a_{15}=62). Exam tip: in pattern questions check whether the difference is constant.

The book numbers are (22,28,34,\ldots) and (S_9=414). Exam tip: take the number of shelves as the number of terms.

The production is the AP (150,175,200,\ldots) and (S_7=1575). Exam tip: treat the daily increase as the common difference.

The distances are (10,12,14,\ldots) and (S_{11}=220). Exam tip: the number of gaps and the number of poles are different.

The marks form the AP (2,5,8,\ldots) and (S_{12}=222). Exam tip: take the number of questions as (n) and find the sum.

Putting (S_n=432) gives (n=16). Exam tip: when total sum is given, form an equation for (n).

This is a decreasing AP and (a_{12}=500+11(-25)=225). Exam tip: write the decrease as a negative difference.

Using (S_8=440) and (d=10) gives (a=20). Exam tip: solve the sum formula for the unknown first term.

The bulb numbers are (45,42,39,\ldots) and (a_8=24). Exam tip: keep the difference negative when moving upward.

The participant numbers are (16,20,24,\ldots) and (S_9=288). Exam tip: treat each round as one term.

The stickers form the AP (5,10,15,\ldots) and (S_{20}=1050). Exam tip: treat page patterns with a constant difference as APs.

Here, the first term is \(a=1000\), the number of terms is \(n=6\), and the sum is \(S_6=9000\). Using the AP sum formula \(S_n=\frac{n}{2}[2a+(n-1)d]\), we get \(9000=\frac{6}{2}[2(1000)+5d]\). Thus, \(9000=6000+15d\), so \(d=200\). Therefore, the monthly increase is 200 rupees. If the increase were 250 rupees, the total for six months would not be 9000 rupees. Exam tip: in questions stating an equal monthly increase, treat that increase as the common difference \(d\).

The seats are (15,19,23,\ldots) and (S_{18}=882). Exam tip: you may also find the last row and use (S_n=\frac{n}{2}(a+l)).

Putting (a_n=700), (100+(n-1)50=700), gives (n=13). Exam tip: use the (n)th term for a particular month's amount.

Using (S_{10}=725) and (a=50) gives (d=5). Exam tip: total work and first day work can give the increase.