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| 1. |
Find the sum of two-digit natural number divisible by 6 |
| Answer» All the natural numbers less than 100 which are divisible by 6 are6, 12, 18, 24,................., 96Here, a1 = 6a2 = 12a3 = 18a4 = 24{tex}\\therefore {/tex}\xa0a2 - a1 = 12 - 6 = 6a3 - a2 = 18 - 12 = 6a4 - a3 = 24 - 18 = 6{tex}{a_2} - {a_1} = {a_3} - {a_2} = {a_4} - {a_3} = 6{/tex}( =6 in each case){tex}\\therefore {/tex}\xa0This sequence is an arithmetic progression whose difference is 6.Here, a = 6d = 6l = 96Let the number of terms be n. Then,l = a + (n - 1)d{tex} \\Rightarrow {/tex}\xa096 = 6 + (n - 1)6{tex} \\Rightarrow {/tex}\xa096 - 6 = (n - 1)6{tex} \\Rightarrow {/tex}\xa090 = (n - 1)6{tex} \\Rightarrow {/tex}\xa0(n - 1)6 = 90{tex} \\Rightarrow n - 1 = \\frac{{90}}{6}{/tex}{tex} \\Rightarrow {/tex}\xa0n - 1 = 15{tex} \\Rightarrow {/tex}\xa0n = 15 + 1{tex} \\Rightarrow {/tex}\xa0n = 16{tex}\\therefore {S_n} = \\frac{n}{2}(a + l){/tex}{tex} = \\left( {\\frac{{16}}{2}} \\right)(6 + 96){/tex}= (8) (102)= 816 | |