A.P.
I solved by first expanding the ap and then eliminating the terms divisible by 3 then i formed 2 different aps (one starting with 8 and the other with 13) each with cd = 15 and summed them up separately. but my answer is wrong whereas the solution given takes the sum of this normal ap first then subtracts it with sum of multiples of 3 in it. i want to understand why my method was wrong?
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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.i could send my solution but its very difficult for me to take a pic since i write vertically in 3 columns to save space (exam practice)
Na it’s good aapne explain. Kiya should work
Why did we make two aps btw?
It is a very clever idea, thoda elaborate kariye
i expanded the original ap a little and eliminated multiples of 3
then i noticed a pattern that every other term is related to a term leaving the next by a cd 15 so that way i wrote two diff aps
like:
8,13,23,28,38,43,....
here, 8 and 23 have cd = 15 and 23 and 38 have cd = 15
so they make one ap
then 13 and 28 have cd = 15 and 28 and 43 have cd = 15
so they make the other ap
so i summed them separately
The 1st,4rth,7th and so on terms are divisible by 3. Toh sum required=(74/2)(373+3)-(121/2)(3+366)
yes i know that approach, it was given in the soln too. i want to understand why my approach is wrong
Your method is absolutely right. You might've made a calculation error or miscalculated the number of terms.
hmm, ok, tho ive checked it
maybe ill retry
thanks for helping
+solved @SirLancelotDuLac
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