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Arithmetic Progressions

All 6 textbook questions with direct answer previews. Open any question for simple explanations and exam-ready answers.

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6
Q1

What is an arithmetic progression? Find the common difference of 45,43,41,39,…45,43,41,39,\ldots.

An arithmetic progression is a list of numbers in which the difference between consecutive terms is constant. For 45,43,41,39,…45,43,41,39,\ldots, common difference d=43−45=−2d=43-45=-2.
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Q2

State the formula for the nnth term of an AP and use it to find the 20th term of 3,7,11,15,…3,7,11,15,\ldots.

For an AP with first term aa and common difference dd, an=a+(n−1)da_n=a+(n-1)d. Here a=3a=3, d=4d=4, so a20=3+19×4=79a_{20}=3+19\times4=79.
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Q3

How many terms are there in the AP 7,10,13,…,437,10,13,\ldots,43?

Here a=7a=7, d=3d=3, and an=43a_n=43. Using an=a+(n−1)da_n=a+(n-1)d, 43=7+(n−1)343=7+(n-1)3, so 36=3(n−1)36=3(n-1), n=13n=13.
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Q4

State the formula for the sum of first nn terms of an AP. Find the sum of first 20 natural numbers.

The sum is Sn=n2[2a+(n−1)d]S_n=\frac n2[2a+(n-1)d] or Sn=n2(a+l)S_n=\frac n2(a+l). For natural numbers, a=1a=1, d=1d=1, n=20n=20, so S20=202(1+20)=210S_{20}=\frac{20}{2}(1+20)=210.
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Q5

Reena starts with a monthly salary of ₹8000 and gets an annual increment of ₹500. What will be her salary in the 10th year?

This is an AP with a=8000a=8000, d=500d=500, and n=10n=10. a10=8000+9×500=12500a_{10}=8000+9\times500=12500.
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Q6

Find the sum of the first 15 terms of the AP 5,8,11,…5,8,11,\ldots.

Here a=5a=5, d=3d=3, n=15n=15. S15=152[2(5)+14(3)]=152(52)=390S_{15}=\frac{15}{2}[2(5)+14(3)]=\frac{15}{2}(52)=390.
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