Question

It is also an Arithmetic progression question please solve..​

Answer 1

Hi !

Solution :- Given first term t1 = 8

t2 = 10 th

so , (common difference) =t2-t1 10-8 = 2

Total term (n) = 60

so, last term tn = a + (n-1)d

tn = 8+ (60-1)2

tn = 8+ 118

tn = 126

last term 'L '= 126

t51= a + (n-1)d

t51 = 8+(51-1)2

t51 = 8+ 100

t51 = 108

t51 is 108

a = 108

Sum of Ap from last term is

sm = m/2(a + L)

s10 = 5(108+126)

s10= 234*5

s10 = 1170 Answer

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Hope it's helpful

Answer 2

Answer:

1170

Step-by-step explanation:

Here,

      First term of this AP is 8

     Common difference between the terms is 12 - 10 = 10 - 8 = 2

From the properties of AP :

• Sum of n terms = ( n / 2 ) [ 2a + ( n - 1 )d ]        ; where a is the first term and d is the common difference between the terms

Here we are said to find the sum of last 10 terms :

So first,

= > 60th term = 8 + ( 60 - 1 )2

= > 60th term = 8 + 59( 2 )

= > 60th term = 8 + 118

= > 60th term = 126

When returning, common difference becomes -ve,

= > S₁₀ = ( 10 / 2 )[ 2( 126 ) + ( 10 - 1 )( -2 ) ]

= > S₁₀ = 5[ 252 - 18 ]

= > S₁₀ = 5( 234 ) = 1170