Numerical answer
In an arithmetic progression, the sum of the first 10 terms is 155 and the sum of the next 10 terms (11th to 20th) is 555. What is the 30th term of the progression?
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Answer: 113.5
Let first term a and common difference d.
S₁₀ = 5(2a + 9d) = 155 ⇒ 2a + 9d = 31.
S₂₀ = 155 + 555 = 710 = 10(2a + 19d) ⇒ 2a + 19d = 71.
Subtracting: 10d = 40 ⇒ d = 4, and 2a = 31 − 36 = −5 ⇒ a = −2.5.
30th term = a + 29d = −2.5 + 116 = 113.5.
Shortcut: consecutive blocks of 10 terms differ by 10 × 10 × d = 100d; here 555 − 155 = 400 ⇒ d = 4 immediately.
S₁₀ = 5(2a + 9d) = 155 ⇒ 2a + 9d = 31.
S₂₀ = 155 + 555 = 710 = 10(2a + 19d) ⇒ 2a + 19d = 71.
Subtracting: 10d = 40 ⇒ d = 4, and 2a = 31 − 36 = −5 ⇒ a = −2.5.
30th term = a + 29d = −2.5 + 116 = 113.5.
Shortcut: consecutive blocks of 10 terms differ by 10 × 10 × d = 100d; here 555 − 155 = 400 ⇒ d = 4 immediately.
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