Question

find the sum of the first 69 terms of an A.P whose 35th term is 30​

Answer 1

Answer:

2070

Step-by-step explanation:

S69 = ?

T35 = 30

let a be the first term,

d be the common difference

$Tn = a + (n - 1)d$

T35 = a + (35 - 1 ) d

30 = a + 34d

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

$Sn = n/2 (2a + (n - 1)d)$

S69 = 69/2 ( 2a + (69 - 1)d)

= 69/2 (2a + 68d)

= 69 × (a + 34d)

= 69 × 30

S69 = 2070.

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Answer 2

Answer:

$s_{69} =$ 4140

Step-by-step explanation:

Let a be the first term and d be the common difference.

We know, $a_{35} = a + 34d$

This means, $a + 34d = 30$             eq. 1

Also, $a69 = a + 68d$                       eq. 2

Now, sum of n terms = n(a + l)

Then, $s_{69} = 69 ( a + a + 68d)$

$s_{69} = 69( 2a + 68d)$

$s_{69} = 69 * 2( a + 34d)$

$s_{69} = 138 * 30$                     [From eq. 1]

$s_{69} = 4140$

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