Question

find the sum of 35 term of an A.P. whose 4th term is 11 and 9th is -4

Answer 1

Step-by-step explanation:

Total number of terms, n = 35

a = ?

d = ?

4th term = 11

9th term = -4

nth term of an AP = a + (n -1) d

Therefore, 4th term = a + (4 - 1) d

=> 11 = a + 3d - (i)

& 9th term = a + (9 - 1) d

=> - 4 = a + 8d - (ii)

Subtracting equation (i) from equation (ii), we get

a + 8d - (a + 3d) = - 4 - 11

=> 5d = - 15

=> d = - 5

Putting value of d = -5 in equation (i), we get

11 = a + 3 × (-5)

=> a = 11 + 15

=> a = 26

Therefore, Reqd. sum

= n/2 {2a + (n - 1) d}

= 35/2 { 2 × 26 + 34 × (-5) }

= 35/2 ( 52 - 170 )

= 35/2 × (- 118)

= 35 × (- 59)

= - 2065 (Ans)

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Answer 2

${ \red{ \tt{Given:}}}$

4th term of A.P is 11

9th term of A.P is -4

${ \red{ \tt{To \: Find:}}}$

Sum of the first 35 term of A.P

${ \red{ \tt{Formula \: Used:}}}$

Sn $= \frac{n}{2} [2a+(n-1)d]$

${ \red{ \tt{Solution:}}}$

$\tt{a + 3d=11... (1)}$

$\tt{a +8d=-4...(2)}$

Subtracting equation 2 from equation 1.

$\tt{a +8d=-4}$

$\tt{a +3d = 11}$

$\tt{5d-15}$

$\green{ \tt{d=-3}}$

Substituting value of d in equation 1,

$a + 3(-3) = 11$

$a-9=11$

${\tt{\green{a = 20}}}$

Value of first 35 terms,

S$\small{35}$$\tt{ = \frac{3}{5} [2 (20)+(351) - 3]}$

S$\small{35}$$\tt{= \frac{35}{2} [40 - 102]}$

S$\small{35}$$\tt{= 35 × [-62] }$

S$\small{35}$$\tt{ = 35 x (-31)}$

S$\small{35}$$\tt{ = -1085}$

${ \boxed {\red{Sum \: of \: first \: 35 \: terms \: is \: 1085}}}$: