Progressions for CSAT: AP, GP and HP Explained

A progression is a sequence of numbers that follows a fixed rule from one term to the next. In an arithmetic progression (AP) you add a constant difference each time; in a geometric progression (GP) you multiply by a constant ratio; in a harmonic progression (HP) the reciprocals of the terms form an AP. In CSAT Paper II, a qualifying paper (roughly 33 percent clears it; the marks do not add to your merit rank), progressions appear both directly (find the sum) and in disguise (bouncing balls, fuel prices, children's ages).

Arithmetic progression: add the same difference

An AP has a first term a and a common difference d: the terms are a, a + d, a + 2d, and so on. The nth term is a + (n minus 1)d, and the sum of the first n terms is n/2 times (2a + (n minus 1)d), which is the same as n/2 times (first term + last term). Two properties do most of the exam work: the mean of an AP equals its middle term (when the count is odd), and the number of terms in a range equals (last minus first)/d plus 1.

Worked example (practice bank). Find the sum of all 2-digit numbers exactly divisible by 3. First term 12, last 99, difference 3. Count: (99 minus 12)/3 plus 1 = 30. Sum = 30/2 times (12 + 99) = 15 times 111 = 1665. The pattern (count) times (average of first and last) solves every such divisibility-sum question.

Worked example (UPSC CSE 2017). Thirteen consecutive 2-digit odd numbers have the property that the mean of the first five is 39. What is the mean of all thirteen? In an AP the mean equals the middle term, so the middle of the first five (the 3rd term) is 39. The numbers are 35, 37, 39, 41, 43, continuing to 13 terms: the middle (7th) term is 39 + 2 times 4 = 47, and the mean of the whole AP equals its middle term, 47, option (a).

Worked example (UPSC CSE 2023). Five children have ages a < b < c < d < e with consecutive ages differing by 2. Can statement 1 alone (eldest is 3 times the youngest) or statement 2 alone (average age 8) determine the youngest's age? The ages are a, a+2, a+4, a+6, a+8. From statement 1: a + 8 = 3a, so a = 4. From statement 2: 5a + 20 = 40, so a = 4. Either statement alone answers it, option (b). Recognising the AP structure turns a data-sufficiency question into one equation.

Geometric progression: multiply by the same ratio

A GP has a first term a and a common ratio r: the terms are a, ar, ar squared, ar cubed, and so on. The nth term is a times r to the power (n minus 1), and the sum of the first n terms is a times (r to the power n minus 1) / (r minus 1). If the ratio satisfies |r| < 1, the terms shrink forever and the infinite sum settles at a / (1 minus r).

Worked example (practice bank). A GP starts at 9 with common ratio 0.5. What is its infinite sum? Since |0.5| < 1, the sum is 9 / (1 minus 0.5) = 9 / 0.5 = 18.

Worked example (practice bank). Find the 8th term of a GP with first term 3 and common ratio 4. The 8th term is 3 times 4 to the power 7 = 3 times 16384 = 49152. For large exponents, multiply in stages rather than computing the full power first.

Worked example (UPSC CSE 2021). A ball is dropped from 1.5 m and bounces back to 4/5 of the previous height each time; it stops bouncing once the previous height is below 50 cm. How many times does it hit the ground? The heights form a GP: 1.5, 1.2, 0.96, 0.768, 0.6144, 0.4915... Each drop ends in a hit, and the ball stops bouncing when the height to drop from falls below 0.5 m, which happens at 0.4915 m. Counting the drops: 1.5 (hit 1), 1.2 (hit 2), 0.96 (hit 3), 0.768 (hit 4), 0.6144 (hit 5); the next bounce reaches only 0.4915 m, so no further bounce. That is 5 hits, option (b).

Harmonic progression and the three means

A sequence is a harmonic progression if the reciprocals of its terms form an arithmetic progression; for example 1, 1/2, 1/3, 1/4. There is no simple closed sum formula for HP, but UPSC tests the three means instead. For two positive numbers, the arithmetic mean (AM) is (a + b)/2, the geometric mean (GM) is the square root of ab, and the harmonic mean (HM) is 2ab/(a + b). They obey AM >= GM >= HM, with equality only when a = b, and GM squared = AM times HM.

CSAT's favourite progression disguises

UPSC often hides a progression inside a story. Three disguises repeat: repeated multiplication or shrinkage (the bouncing ball above) signals a GP; prices or ages changing linearly signal an AP or a pair of linear equations; and averages of evenly spaced values let you read the middle term directly.

Worked example (UPSC CSE 2021). From January 1, 2021, petrol on day m costs 80 + 0.1m (m up to 100, then constant) and diesel on day n costs 69 + 0.15n. On which date are the prices equal? Set 80 + 0.1d = 69 + 0.15d: then 11 = 0.05d, so d = 220. But petrol stops rising after day 100, when it is 80 + 10 = 90. So solve 69 + 0.15d = 90: 0.15d = 21, d = 140. Day 140 of 2021: January (31) + February (28) + March (31) + April (30) = 120, plus 20 days into May = 20th May, option (b). The trap is solving only the first equation and ignoring the cap at day 100.

Formula cheat sheet

Traps UPSC sets

• The fencepost error. Numbers from 12 to 99 divisible by 3 are 30 in count, not 29: always add 1 after dividing the span by the step.
• Mean is the middle only for symmetric sequences. The 39 trick works because the numbers are consecutive odds; do not apply it to arbitrary lists.
• Infinite sum needs |r| < 1. With ratio 2 the series explodes; the formula a/(1 minus r) applies only to shrinking ratios like 0.5 or 4/5.
• Caps and ceilings in stories. The petrol price stops rising at day 100; solving the linear equation alone gives the wrong day 220.
• Stopping conditions in bounce problems. The ball hits the ground one more time than it bounces; count drops (hits), not bounces, and check the threshold against the height to drop from.

Speed tactics

• Sum = count times average. For any evenly spaced range, sum = number of terms times (first + last)/2; you never need the long AP sum formula.
• Odd-count mean = middle term. Thirteen consecutive odds with middle 47: the mean is 47, no addition needed.
• List GP heights until the threshold. Bounce problems have at most six or seven terms; write them out (1.5, 1.2, 0.96...) instead of solving inequalities.
• Read the disguise. Shrinkage by a fixed fraction is a GP; equal steps are an AP; if you spot the type in ten seconds, the formula does the rest.

Key Terms

• Sequence: an ordered list of numbers; each position in the list holds one term.
• Series: the sum of the terms of a sequence, such as the sum of the first n terms of an AP.
• Term: a single number in a sequence, identified by its position: first term, nth term, and so on.
• Common difference (d): the constant added to each term to get the next one in an arithmetic progression.
• Common ratio (r): the constant multiplied with each term to get the next one in a geometric progression.
• Nth term: a formula giving the term at position n directly, without listing all previous terms.
• Partial sum: the sum of the first n terms of a sequence, written S sub n.
• Infinite sum (convergence): the limit a shrinking series approaches; it exists for a GP only when |r| < 1.
• Arithmetic mean: the ordinary average (a + b)/2; the middle term of a three-term AP.
• Geometric mean: the square root of ab; the middle term of a three-term GP.
• Harmonic mean: 2ab/(a + b); the reciprocal of the arithmetic mean of the reciprocals, used for average speeds over equal distances.

Practice questions