-
Step : 1
solve:-
Term at 5 position is 3 *3(5-1)
= 3 *3(4)
= 3 * 81
= 243
-
Step : 2
Now,
The sum of 5 to 9 position is 243*[1 - 3(5)] /[1-3]
= (243)[ 1 - 3(5)]/[1-3]
= (243)(-242)/(-2)
= (243)(121)
ie. S5 to S9 = 29403
-
Step : 3
Second method:-
Term at 5 position is 3 *3(5-1)
= 3 *3(4)
= 3 * 81
= 243 (=t1)
-
Step : 4
Term at 6 position is 3 *3(6-1)
= 3 *3(5)
= 3 * 243
= 729 (=t2)
-
Step : 5
Term at 7 position is 3 *3(7-1)
= 3 *3(6)
= 3 * 729
= 2187 (=t3)
-
Step : 6
Term at 8 position is 3 *3(8-1)
= 3 *3(7)
= 3 * 2187
= 6561 (=t4)
-
Step : 7
Term at 9 position is 3 *3(9-1)
= 3 *3(8)
= 3 * 6561
= 19683 (=t5)
-
Step : 8
so the sequance of numbers from position 5 to 9 are 243, 729, 2187, 6561, 19683
= 243 + 729 + 2187 + 6561 + 19683;
= 29403
Calculation Speed Booster
✈ Click on the 《more so》 button in the timer panel .
✈Another question will appear with different numerical values .
✈Now try to solve it faster.
✈Check your speed by the timer.
✈Share your speed on
✈Share your speed on
✈
JOIN our TELEGRAM group : @AptitudeMathSpeedBoosterEatMaths
.
Want some question of your choice
If you want to practice some questions which are not available here then send us the details of the question in our whatsapp 7291934297.
Want to practice on hundreds of varity of such questions
Call 7463918936, and book your limited premium membership. Monthly rupees 150/- only. Online tutorial classes also available seprately for making your basics and theory strong.
