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.
