7Pairs from the endsTk+Tn+1−k=a+lkth from the start+kth from the end
8Equal sums cancelSm=Sn⇒Sm+n=0m=n counts of terms from the start
9Blocks of k termsS2k−2Sk=k2deach next block of k terms is k2d more
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
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Q1Easy
How many terms does the arithmetic progression 7,11,15,…,199 have?
Correct answer·49
Why
Count the jumps, then add one for the first term.
From 7 to 199 is 192÷4=48 jumps of 4.
So the progression has 48+1=49 terms.
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 answer·780
Why
Two known terms give the step at once: d=12−545−17=4.
Walk back four steps to get T1=17−16=1, and forward to get T20=1+76=77.
Count times the average of the ends: 20×39, which is 780.
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