Question

solve this ap problem

Answer 1

a10 = -37
S6= -27
a10= a+9d
-37= a+9d
S6=n/2(2a+(n-1)d) 
-27=3(2a+5d)
-9=2a + 5d




(-37= a+9d ) 2                     -74=2a+18d
(-9=2a + 5d)⇒⇒⇒⇒⇒        (-9=2a + 5d)
                                             +__-___-_________
                                             -65=13d
                                              d=-5
                                              a=8

_____________Thus use an=a+(n-1)d formula and get 8 terms

Answer 2

Tenth Term of an Arithmetic Progression = $t_{10}$ = -37
Sum of the first six terms = $S_{6}$ = -27

Let a denote the notation for the first term
Let d denoted the notation for common difference

Now, $t_{10} = -37$
$-37 = a + (10-1)d$ ... Using formula for n(th) term[
$a + 9d = -37$

In the same way, find out the Equation for 
$S_{6} = -27$
$-27 = \frac{6}{2} [2a + (6-1)d]$
$-27 = 3 [2a + 5d]$
$\frac{-27}{3} = 2a +5d$ Moving 3 to the LHS
$-9 = 2a + 5d$
$2a + 5d = -9$ ... eq (2)

Multiplying eq. (1) by 2 we get,
$2a + 18d = -74$ ... eq (3)

Subtracting eq. (2) from (3), we get,
$2a + 18d = -74$
$- 2a + 5d = -9$

$13d = -65$
$d = \frac{-65}{13} = -5$

Now that we have got the value of d (common difference) we can calculate the value of "a"

By putting d = -5 in eq. (1) we get,
a + 9 (-5) = -37
a - 45 = -37 
Therefore, a = -37 + 45 = 8

Now that we have got the value of a and d,
$S_{8} = \frac{8}{2} [ 2(8) + (8-1) -5 ]$
$S_{8} = 4 [16 + 7 (-5) ]$
$S_{8} = 4 [19]$
$S_{8} = -19 X 4 = -76$

Therefore, the sum of the first 8 terms of an A.P is -76