• Q. First term is of arithmatic progression sequence is 32 and the difference between consecutive numbers is 6 , Find the sum numbers 36th, 37th, 38th, 39th and 40th terms.
    Share to Whatsapp



  • Sum of 36 position is (36/2)[2*32 + (36 - 1)x 6 ]

    = (18)[64 + (35)x 6 ]

    = (18)[64 + 210 ]

    = (18)[274]

    ie. S36 = (4932)

  • Sum of 37 position is (37/2)[2*32 + (37 - 1)x 6 ]

    = (18.5)[64 + (36)x 6 ]

    = (18.5)[64 + 216 ]

    = (18.5)[280]

    ie. S37 = (5180)

  • Sum of 38 position is (38/2)[2*32 + (38 - 1)x 6 ]

    = (19)[64 + (37)x 6 ]

    = (19)[64 + 222 ]

    = (19)[286]

    ie. S38 = (5434)

  • Sum of 39 position is (39/2)[2*32 + (39 - 1)x 6 ]

    = (19.5)[64 + (38)x 6 ]

    = (19.5)[64 + 228 ]

    = (19.5)[292]

    ie. S39 = (5694)

  • Sum of 40 position is (40/2)[2*32 + (40 - 1)x 6 ]

    = (20)[64 + (39)x 6 ]

    = (20)[64 + 234 ]

    = (20)[298]

    ie. S40 = (5960)

  • so the sequance of the sum numbers from position are.
    S36 = 4932
    S37 = 5180
    S38 = 5434
    S39 = 5694
    S40 = 5960

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 onShare to Whatsapp

✈Share your speed on Share to Telegram

✈ 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.