Question

find the sum of the AP : 37,33,29...... to 12 terms​

Answer 1

Answer:

a=37

a=37d=-4

a=37d=-4n=12

a=37d=-4n=12Sn = n/2[2a+(n-1)d]

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]= 6[74-44)

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]= 6[74-44)= 6[30]

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]= 6[74-44)= 6[30]= 180

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]= 6[74-44)= 6[30]= 180SUM OF AP TO 12 TERMS IS 180.

a=37d=-4n=12Sn = n/2[2a+(n-1)d]= 12/2[37×2+11×(-4)]= 6[74-44)= 6[30]= 180SUM OF AP TO 12 TERMS IS 180.KINDLY MARK AS BRAINLIEST