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The numbers lying between 107 and 253 that are multiples of 5 are 110, 115, 120,…, 250 a2 – a1 = 115 – 110 = 5 a3 – a2 = 120 – 115 = 5 ∵ a3 – a2 = a2 – a1 = 5 Therefore, the series is in AP Here, a = 110, d = 5 and an = 250 We know that, an = a + (n – 1)d ⇒ 250 = 110 + (n – 1)5 ⇒ 250 – 110 = (n – 1)5 ⇒ 140 = (n – 1)5 ⇒ 28 = (n – 1) ⇒ n = 29 Now, we have to find the sum of this AP ⇒ S29 = 29[110 + 14 × 5] ⇒ S29 = 29[180] ⇒ S29 = 5220 Hence, the sum of all numbers lying between 107 and 253 is 5220.
The answer is 5220.
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