Find the number of terms which is common to two arithmetic progressions: 3,7,11,. 407 and 2,9,16,. 709.

The first sequence increments by 4 with each step, and the second increments by 7. From this you know that when a common number occurs the next highest common number is 4 *7 = 28 higher. So first find the lowest number that's common to both sets, and then figure out how many times you can increment that by 28 before you exceed 407. There's no point in looking above 407 because no numbers greater than that are in the first set. Make sense? Post back and tell us what you find.

First you have to find the common diference thenuse tn formula and with the answer you get divide it with 3.that shouls do.