the sum of the first n terms of an ap is SN is equal to 3n² + 5n find its first term and common difference and 25th term and sum of the 25th to 35 the terms

Answer: First term a=8a = 8, common difference d=6d = 6, 25th25^{\text{th}} term a25=152a_{25} = 152, and the sum of the 25th25^{\text{th}} to 35th35^{\text{th}} terms is 20022002.

Step-by-step solution

Step 1: Find the first term aa

The sum of the first 1 term of an arithmetic progression is equal to the first term itself, so a1=S1a_1 = S_1. Substituting n=1n = 1 into the given formula Sn=3n2+5nS_n = 3n^2 + 5n yields S1=8S_1 = 8. Therefore, the first term aa is 8.

Step 2: Find the second term and common difference dd

To determine the common difference, we first calculate S2S_2 by substituting n=2n = 2 into the sum formula, giving S2=22S_2 = 22. The second term is a2=S2−S1=22−8=14a_2 = S_2 - S_1 = 22 - 8 = 14. The common difference is given by d=a2−a1=14−8=6d = a_2 - a_1 = 14 - 8 = 6.

Step 3: Find the 25th25^{\text{th}} term

We use the standard nthn^{\text{th}} term formula for an AP, an=a+(n−1)da_n = a + (n - 1)d. Substituting a=8a = 8, d=6d = 6, and n=25n = 25, we get a25=8+24×6=152a_{25} = 8 + 24 \times 6 = 152.

Step 4: Calculate the sum of terms from 25th25^{\text{th}} to 35th35^{\text{th}}

The sum of terms from the 25th25^{\text{th}} term to the 35th35^{\text{th}} term inclusive is obtained by subtracting the sum of the first 24 terms from the sum of the first 35 terms, namely S35−S24S_{35} - S_{24}. We compute S35=3(1225)+175=3850S_{35} = 3(1225) + 175 = 3850 and S24=3(576)+120=1848S_{24} = 3(576) + 120 = 1848. Subtracting these yields 3850−1848=20023850 - 1848 = 2002.

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