Given a1=28,an=-77 and Sn=-196 in an A.P, find the 10th term and the sum of the first 10 terms.
whats ur answer?
a1 = 28 , an = - 77 , Sn = - 196
Sn = ( n / 2 ) ( a1 + an )
- 196 = ( n / 2 ) ( 28 - 77 )
- 196 = ( n / 2 ) ∙ ( - 49 )
Multiply both sides by - 2
392 = 49 n
n = 392 / 49
n = 8
an = a8 = - 77
a8 = - 77
an = a1 + ( n - 1 ) d
a8 = - 77 = 28 + ( 8 - 1 ) ∙ d
- 77 = 28 + 7 d
- 77 - 28 = 7 d
- 105 = 7 d
d = - 105 / 7 = - 15
a10 = 28 + ( 10 - 1 ) ∙ ( - 15 )
a10 = 28 + 9 ∙ ( - 15 )
a10 = 28 - 135
a10 = - 107
To find the 10th term and the sum of the first 10 terms, we need to use the formulas for an arithmetic progression (A.P).
The formula for the nth term (an) of an A.P is:
an = a1 + (n-1)d
where 'a1' is the first term of the A.P, 'n' is the term number, and 'd' is the common difference between the terms.
In this case, given that a1 = 28 and an = -77, we can substitute these values into the formula and solve for 'd'.
-77 = 28 + (n-1)d
-105 = (n-1)d
Next, we have the formula for the sum of the first 'n' terms (Sn) of an A.P:
Sn = (n/2)[2a1 + (n-1)d]
To find the common difference 'd' and substitute it into this formula to find the sum Sn.
Using the given information, Sn = -196 and n = 10. We can substitute these values into the formula and simplify the equation to solve for 'd'.
-196 = (10/2)[2(28) + (10-1)d]
-196 = 5(56 + 9d)
-196 = 280 + 45d
-476 = 45d
d = -10.6
Now, we have the value of the common difference 'd' as -10.6. We can substitute this value back into the equation for the nth term (an) to find the 10th term.
a10 = a1 + (10-1)d
a10 = 28 + 9(-10.6)
a10 = 28 - 95.4
a10 = -67.4
Therefore, the 10th term of the arithmetic progression is -67.4.
To find the sum of the first 10 terms (S10), we substitute the known values into the formula for Sn:
S10 = (10/2)[2(28) + (10-1)(-10.6)]
S10 = 5[56 + 9(-10.6)]
S10 = 5[56 - 95.4]
S10 = 5(-39.4)
S10 = -197
Therefore, the sum of the first 10 terms of the arithmetic progression is -197.