Arithmetic Progression

An arithmetic progression is fully defined by a first term a and a common difference d, with the nth term given by a + (n-1)d and the sum of n terms by S_n = (n/2)[2a + (n-1)d]. This is a standard JEE Advanced linear-system setup. From the seventh term, a + 6d = 34, and from the fifteenth term, a + 14d = 74. Subtracting the first from the second gives 8d = 40, so d = 5, and back-substitution gives a + 30 = 34, hence a = 4. The sum of twenty terms is S_20 = (20/2)[2(4) + 19(5)] = 10[8 + 95] = 10 × 103 = 1030. Option 1050 results from using a = 5 by mis-solving the system. Option 990 arises from taking d = 4 instead of 5. Option 1130 comes from adding rather than subtracting the equations. The reliable strategy whenever two term conditions are given is to treat them as a two-by-two linear system in a and d, since eliminating one unknown by subtraction immediately exposes the common difference, after which back-substitution recovers the first term cleanly. Plausibility check: the average of the first and twentieth terms is (4 + 99)/2 = 51.5, and 51.5 × 20 = 1030, confirming consistency with the average-term interpretation of an AP sum.