Skip to content
Top 100 Most Asked QuestionsLaunching soon20% off the yearly Test Series pass
Klay
Start Learning

LearnQuantitative AptitudeAlgebra

03.11 · Algebra

Arithmetic Progression

Elite lesson10 formulas10 questions

On the syllabus of

Elite lesson22 minEnglish

Formulas

At a glance 10 formulas

  1. 1nth termTn=a+(n−1) dT_n = a + (n - 1)\,da=first term, d=common differencea = \text{first term},\ d = \text{common difference}
  2. 2Number of termsn=l−ad+1n = \dfrac{l - a}{d} + 1l=last term; the +1 counts the first terml = \text{last term; the } {+}1 \text{ counts the first term}
  3. 3Sum of an APSn=n2 (a+l)S_n = \dfrac{n}{2}\,(a + l)Sn=n2 [2a+(n−1)d]S_n = \dfrac{n}{2}\,[2a + (n - 1)d]
  4. 4Three in a rowb=a+c2b = \dfrac{a + c}{2}a,b,c consecutive; in a GP: b2=aca, b, c \text{ consecutive; in a GP: } b^2 = ac
  5. 5Term from the sumTn=Sn−Sn−1T_n = S_n - S_{n-1}Sn=pn2+qn⇒T1=p+q, d=2pS_n = pn^2 + qn \Rightarrow T_1 = p + q,\ d = 2p
  6. 6Odd count: middle termS2k−1=(2k−1) TkS_{2k-1} = (2k - 1)\,T_ke.g. S5=5 T3,  S9=9 T5\text{e.g. } S_5 = 5\,T_3,\ \ S_9 = 9\,T_5
  7. 7Pairs from the endsTk+Tn+1−k=a+lT_k + T_{n+1-k} = a + lkth from the start+kth from the endk\text{th from the start} + k\text{th from the end}
  8. 8Equal sums cancelSm=Sn⇒Sm+n=0S_m = S_n \Rightarrow S_{m+n} = 0m≠n counts of terms from the startm \ne n \text{ counts of terms from the start}
  9. 9Blocks of k termsS2k−2Sk=k2dS_{2k} - 2S_k = k^2 deach next block of k terms is k2d more\text{each next block of } k \text{ terms is } k^2 d \text{ more}
  10. 10Common termsThe shared terms form a new AP: start at the first number in both lists and step by the LCM of the two common differences, up to the smaller last term.

Test yourself

10 questions · 3 easy · 4 medium · 3 hard

Tap an option to answer. The first tap counts.

  1. Q1Easy

    How many terms does the arithmetic progression 7,11,15,…,1997, 11, 15, \dots, 199 have?

    Correct answer49

    Why

    Count the jumps, then add one for the first term.

    From 7 to 199 is 192÷4=48192 \div 4 = 48 jumps of 4.

    So the progression has 48+1=4948 + 1 = 49 terms.

  2. Q2Easy

    The 5th term of an arithmetic progression is 17 and its 12th term is 45. What is the sum of its first 20 terms?

    Correct answer780

    Why

    Two known terms give the step at once: d=45−1712−5=4d = \frac{45 - 17}{12 - 5} = 4.

    Walk back four steps to get T1=17−16=1T_1 = 17 - 16 = 1, and forward to get T20=1+76=77T_{20} = 1 + 76 = 77.

    Count times the average of the ends: 20×3920 \times 39, which is 780.

8 more questions: 1 easy, 4 medium, 3 hard

Sign in free to see all ten and get your score. No payment, no card.

More practice: IPMAT, IIMB, FEAT, MUSAT, AAA, MSAT, NPAT, SET and Christ BBA previous year questions

Logos and exam names belong to their owners. Klay Prep is not affiliated with them.