Question

the first two terms of an arithmetic progression are 16 and 24 find the least number of terms of the progression which must be taken for their sum to exceed 20 000

Answer 1

Answer:

Sum exceeds 20000 for 70 terms

Step-by-step explanation:

Given two terms of AP are 16 and 24

First term, a= 16 ; Common difference, d = 24–16 =8

Let n be the number of terms such that their sum Sn > 20,000

Sn = (n/2)[2a+(n-2)d]

Let Sn = 20000

20000 = (n/2)[2(16)+(n-1)8]

n²+3n-5000 = 0

Taking only positive root, n = 69.226

n≈70

For n= 69 , Sn = 19872 < 20000

For n= 70, Sn = 20440 > 20000

Ans: Sum exceeds 20000 for 70 terms

Answer 2

Answer:

Mark me as brainlist

Step-by-step explanation:

Given two terms of AP are 16 and 24

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 20000

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]n²+3n-5000 = 0

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]n²+3n-5000 = 0Taking only positive root, n = 69.226

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]n²+3n-5000 = 0Taking only positive root, n = 69.226n≈70

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]n²+3n-5000 = 0Taking only positive root, n = 69.226n≈70For n= 69 , Sn = 19872 < 20000

Given two terms of AP are 16 and 24First term, a= 16 ; Common difference, d = 24–16 =8Let n be the number of terms such that their sum Sn > 20,000Sn = (n/2)[2a+(n-2)d]Let Sn = 2000020000 = (n/2)[2(16)+(n-1)8]n²+3n-5000 = 0Taking only positive root, n = 69.226n≈70For n= 69 , Sn = 19872 < 20000For n= 70, Sn = 20440 > 20000

Ans: Sum exceeds 20000 for 70 terms